General Chemistry

The following practice problems are to master to topics on the ideal gas laws: Boyle’s law, Charles’s law, and Avogadro’s Law, as well as the combined gas law equation. There are examples to work on the Dalton law of partial pressures, the Graham’s law of effusion, and gas stoichiometry.

Here are the links to the topics covered in this practice set:

Practice

1.

The pressure of a gas is 2.30 atm in a 1.80 L container. Calculate the final pressure of the gas if the volume is decreased to 1.20 liters.

3.45 atm

Solution

First, write down what you gave and what needs to be determined:

P1 = 2.30 atm

V1 = 1.80 L

V2 = 1.20 L

P2 = ?

Next, write the combined gas law equating and get rid of the constants. In this case, the temperature and the moles of the gas are constant, so by eliminating them, we get the Boyle’s laws:

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}$

${{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}\; = \;{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}$

Now, rearrange to calculate P2:

${{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}$

${{\rm{P}}_{\rm{2}}}\; = \;\frac{{{\rm{2}}{\rm{.30}}\;{\rm{atm}}\;{\rm{ \times 1}}{\rm{.80}}\;\cancel{{\rm{L}}}}}{{{\rm{1}}{\rm{.20}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.45}}\;{\rm{atm}}$

video

2.

After changing the pressure of a gas sample from 760.0 torr to 0.800 atm, it occupies 4.30 L volume. What was the initial volume of the gas?

3.44 L

Solution

First, write down what you gave and what needs to be determined:

P1 = 760.0 torr

V1 = ?

V2 = 4.30 L

P2 = 0.800 atm

Next, write the combined gas law equating and get rid of the constants. In this case, the temperature and the moles of the gas are constant, so by eliminating them, we get the Boyle’s laws:

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}$

${{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}\; = \;{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}$

Now, rearrange to calculate V1:

${{\rm{V}}_{\rm{1}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{P}}_{\rm{1}}}}}$

One last thing is to convert the initial and final pressure to the same units. It does not matter which one, so let’s convert torr to atm:

${{\rm{P}}_{\rm{1}}}\; = \;{\rm{760}}{\rm{.0}}\;\cancel{{{\rm{torr}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}{\rm{.0}}\;\cancel{{{\rm{torr}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.000}}\;{\rm{atm}}$

Therefore,

${{\rm{V}}_{\rm{1}}}\; = \;\frac{{{\rm{0}}{\rm{.800}}\;\cancel{{{\rm{atm}}}}\;{\rm{ \times 4}}{\rm{.30}}\;{\rm{L}}}}{{{\rm{1}}\;\cancel{{{\rm{atm}}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.44}}\;{\rm{L}}$

3.

What will be the final volume of a 3.50 L sample of nitrogen at 20.0 °C if it is heated to 200. °C?

5.65 L

Solution

First, write down what you gave and what needs to be determined:

T1 = 20.0 oC

V1 = 3.50 L

T2 = 200. oC

V2 = ?

Remember, the temperature must always be converted to K in gas problems, so the first thing here is to do the conversions:

T1 = 20.0 + 273 = 293 K

T2 = 200. + 273 = 473 K

Next, write the combined gas law equating and get rid of the constants. In this case, the pressure and the moles of the gas are constant, so by eliminating them, we get the Charles’s law:

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}$

$\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}$

Now, rearrange to calculate V2:

${{\rm{V}}_{\rm{2}}}\; = \;\frac{{{{\rm{T}}_{\rm{2}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}$

${{\rm{V}}_{\rm{2}}}\; = \;\frac{{{\rm{473}}\;\cancel{{\rm{K}}}\;{\rm{ \times }}\;{\rm{3}}{\rm{.50}}\;{\rm{L}}}}{{{\rm{293}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.65}}\;{\rm{L}}$

So, the volume is increasing from 3.50 L to 5.65 L which makes sense because the temperature is increased and the gas is expanding.

video

4.

The volume of a gas decreased from 2.40 L to 830. mL and the final temperature is set at 40.0 °C. Assuming a constant pressure, calculate the initial temperature of the gas in kelvins.

905 K

Solution

First, write down what you gave and what needs to be determined:

V1 = 2.40 L

T2 = 40.0 oC

V2 = 830. mL

T1 = ? oC

Remember, the temperature must always be converted to K in gas problems, so the first thing here is to do the conversions:

T2 = 40.0 + 273 = 313 K

The volumes are also in different units, so we can convert V2 to L.

V2 = 830. mL x 1 L/1000 mL = 0.830 L

Next, write the combined gas law equating and get rid of the constants. In this case, the pressure and the moles of the gas are constant, so by eliminating them, we get the Charles’s law:

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}$

$\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}$

Now, rearrange to calculate T1:

${{\rm{T}}_{\rm{1}}}\; = \;\frac{{{{\rm{T}}_{\rm{2}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}$

${{\rm{T}}_{\rm{1}}}\; = \;\frac{{{\rm{313}}\;{\rm{K}}\;{\rm{ \times }}\;{\rm{2}}{\rm{.40}}\;\cancel{{\rm{L}}}}}{{{\rm{0}}{\rm{.830}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{905}}\;{\rm{K}}$

5.

A sample of helium gas at 1.40 atm is heated from 23.0 °C to 400.0 K. How many atmospheres is the final pressure of the helium gas?

1.89 atm

Solution

First, write down what you gave and what needs to be determined:

P1 = 1.40 atm

T1 = 23.0 oC

T2 = 400.0 K

P2 = ?

Remember, the temperature must always be converted to K in gas problems, so the first thing here is to do the conversions:

T1 = 23.0 + 273 = 296 K

Next, write the combined gas law equating and get rid of the constants. In this case, the pressure and the moles of the gas are constant, so by eliminating them, we get the Gay-Lussac’s Law:

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}$

$\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}$

Rearrange the equation to find P2:

${{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{\rm{1}}{\rm{.40}}\;{\rm{atm}}\;{\rm{ \times }}\;{\rm{400}}{\rm{.0}}\;\cancel{{\rm{K}}}}}{{{\rm{296}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.89}}\;{\rm{atm}}$

video

6.

A sample of hydrogen gas is added in a 5.80 L container at 56.0 °C. How many moles of the gas are present in the container if the pressure is 6.70 atm?

1.44 mol

Solution

Let’s write down what we are given:

V = 5.80 L

T = 56.0 oC

P = 6.70 atm

n(H2) = ?

T = 273 + 56.0 = 329 K

To find the moles of hydrogen, we need the ideal gas law equation:

PV = nRT

${\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{PV}}}}{{{\rm{RT}}}}$

${\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{6}}{\rm{.70}}\;\cancel{{{\rm{atm}}}}\; \times \;{\rm{5}}{\rm{.80}}\;\cancel{{\rm{L}}}}}{{{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}{\rm{ }} \times \;{\rm{329}}\;\cancel{{\rm{K}}}}}\; = \;1.44\;{\rm{mol}}$

7.

What is the pressure in a 26.0 L container with 5.40 moles of nitrogen dioxide if the temperature is 64.0°C?

5.74 atm

Solution

Let’s write down what we are given:

V = 26.0 L

n(NO2) = 5.40 mol

T = 64.0 oC

P = ?

T = 273 + 64.0 = 337 K

To find the pressure, we need the ideal gas law equation:

PV = nRT

${\rm{P}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{V}}}$

${\rm{P}}\;{\rm{ = }}\;\frac{{{\rm{5}}{\rm{.40}}\;\cancel{{{\rm{mol}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;{\rm{atm}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}\;{\rm{ \times }}\;{\rm{337}}\;{\rm{K}}}}{{{\rm{26}}{\rm{.0 }}\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.74}}\;{\rm{atm}}$

8.

A 3.7 L gas sample, initially at STP, is heated to 280. °C at constant volume. Calculate the final pressure of the gas in atm.

2.03 atm

Solution

First, write down what you gave and what needs to be determined:

V1 = V2 = 3.7 L

P1 = 1 atm

T1 = 0 oC

T2 = 280. oC

P2 = ?

Remember, the temperature must always be converted to K in gas problems, so the first thing here is to do the conversions:

T1 = 0 + 273 = 273 K

T2 = 280. +273 = 553 K

Next, write the combined gas law equating and get rid of the constants. In this case, the pressure and the moles of the gas are constant, so by eliminating them, we get the Gay-Lussac’s Law:

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}$

$\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}$

Rearrange the equation to find P2:

${{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{\rm{1}}{\rm{.00}}\;{\rm{atm}}\;{\rm{ \times }}\;{\rm{553}}\;\cancel{{\rm{K}}}}}{{{\rm{273}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.03}}\;{\rm{atm}}$

9.

A 2.65 g sample of dry ice (solid carbon dioxide) is placed in a 2.90-L vessel and converted into CO2 gas. Calculate the pressure inside the vessel if the temperature is at 35.0 oC.

0.525 atm

Solution

Let’s write down what we are given:

m(CO2) = 2.65 g

V = 2.90 L

T = 35.0 oC

P = ?

T = 273 + 35.0 = 308 K

To find the pressure, we need the ideal gas law equation:

PV = nRT

${\rm{P}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{V}}}$

One thing still missing in the equation is the moles of CO2, which can be calculated from its mass:

n(CO2) = 2.65 g/44.0 g/ mol = 0.0602 mol

${\rm{P}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.0602}}\;\cancel{{{\rm{mol}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;{\rm{atm}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}\;{\rm{ \times }}\;{\rm{308}}\;\cancel{{\rm{K}}}}}{{{\rm{2}}{\rm{.90 }}\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.525}}\;{\rm{atm}}$

10.

A gas sample is stored in a 429 mL container at 9.50°C and 2.20 atm. Calculate the pressure of the gas if the volume changes to 134 mL and the container is heated to 134.5°C? Assume a constant amount of gas.

10.2 atm

Solution

First, write down what you gave and what needs to be determined:

V1 = 429 mL

V2 = 134 mL

P1 = 2.20 atm

T1 = 9.50 oC

T2 = 134.5 oC

P2 = ?

Remember, the temperature must always be converted to K in gas problems, so the first thing here is to do the conversions:

T1 = 9.50 + 273 = 282.5 K

T2 = 134.5 +273 = 407.5 K

Next, write the combined gas law equating and get rid of the constants. In this case, only the moles of the gas are constant:

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}$

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}$

${{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}{{\rm{V}}_{\rm{2}}}}}$

${{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}{{\rm{V}}_{\rm{2}}}}}\; = \;\frac{{{\rm{2}}{\rm{.20}}\;{\rm{atm}}\;{\rm{ \times }}\;{\rm{429}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;{\rm{407}}{\rm{.5}}\;\cancel{{\rm{K}}}}}{{{\rm{282}}{\rm{.5}}\;\cancel{{{\rm{K}}\;}}\;{\rm{134}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{10}}{\rm{.2}}\;{\rm{atm}}$

Notice that we did not need to convert the ml to L, because they cancel out in the equation. The volume must be in L though when the R, expressed using L, is part of the calculation. Remember also that temperature must be converted to K regardless of the equation. The difference between V and T is that for the volume, we multiply or divide by 1000 when converting between ml and L, while for temperature, we add 273 which does not change both numbers with an equal magnitude.

11.

A gas sample occupies 22.0 L at 171°C and 1.43 atm. Calculate the volume of the gas if its temperature and pressure are increased to 197 °C and 1.80 atm respectively.

18.5 L

Solution

First, write down what you gave and what needs to be determined:

V1 = 22.0 L

P1 = 1.43 atm

T1 = 171 oC

T2 = 197 oC

P2 = 1.80 atm

V2 = ?

Remember, the temperature must always be converted to K in gas problems, so the first thing here is to do the conversions:

T1 = 171 + 273 = 444 K

T2 = 197 +273 = 470 K

Next, write the combined gas law equating and get rid of the constants. In this case, only the moles of the gas are constant:

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}$

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}$

${{\rm{V}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}{{\rm{P}}_{\rm{2}}}}}\; = \;\frac{{{\rm{1}}{\rm{.43}}\;\cancel{{{\rm{atm}}}}\;{\rm{ \times }}\;{\rm{22}}{\rm{.0}}\;{\rm{L}}\;{\rm{ \times }}\;{\rm{470}}\cancel{{\rm{K}}}}}{{{\rm{444}}\cancel{{{\rm{K}}\;}}\;{\rm{1}}{\rm{.80}}\;\cancel{{{\rm{atm}}}}}}\;{\rm{ = }}\;{\rm{18}}{\rm{.5}}\;{\rm{L}}$

12.

A sample of ethylene gas (C2H6) collected in a 36.4 mL vessel with a freely moving piston, at 31.0 °C exerts 745 mmHg pressure. What is the volume of this gas at STP?

32.0 mL

Solution

First, write down what you gave and what needs to be determined:

V1 = 36.4 mL

P1 = 745 mmHg

T1 = 31.0 oC

T2 = 0 oC

P2 = 1.00 atm

V2 = ?

Remember, the temperature must always be converted to K in gas problems, so the first thing here is to do the conversions:

T1 = 31.0 + 273 = 304 K

T2 = 0 +273 = 273 K

Convert the mmHg to atm as well:

P1 = 745 mmHg x 1 atm/760 mmHg = 0.980 atm

Next, write the combined gas law equating and get rid of the constants. In this case, only the moles of the gas are constant:

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}$

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}$

${{\rm{V}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}{{\rm{P}}_{\rm{2}}}}}\; = \;\frac{{{\rm{0}}{\rm{.980}}\;\cancel{{{\rm{atm}}}}\;{\rm{ \times }}\;{\rm{36}}{\rm{.4}}\;{\rm{mL}}\;{\rm{ \times }}\;{\rm{273}}\cancel{{\rm{K}}}}}{{{\rm{304}}\cancel{{{\rm{K}}\;}}\;{\rm{1}}{\rm{.00}}\;\cancel{{{\rm{atm}}}}}}\;{\rm{ = }}\;{\rm{32}}{\rm{.0}}\;{\rm{mL}}$

13.

A gas-filled balloon having a volume of 3.50 L at 1.30 atm and 25.0 °C is allowed to rise 5 km above the surface of Earth, where the temperature and pressure are 12.0 °C and 1.10 atm, respectively. What would the volume of the balloon be in these conditions?

3.96 L

Solution

First, write down what you gave and what needs to be determined:

V1 = 3.50 L

P1 = 1.30 atm

T1 = 25.0 oC

T2 = 12.0 oC

P2 = 1.10 atm

V2 = ?

Remember, the temperature must always be converted to K in gas problems, so the first thing here is to do the conversions:

T1 = 25.0 + 273 = 298 K

T2 = 12.0 +273 = 285 K

Next, write the combined gas law equating and get rid of the constants. In this case, only the moles of the gas are constant:

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}$

$\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}$

${{\rm{V}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}{{\rm{P}}_{\rm{2}}}}}\; = \;\frac{{{\rm{1}}{\rm{.30}}\;\cancel{{{\rm{atm}}}}\;{\rm{ \times }}\;{\rm{3}}{\rm{.50}}\;{\rm{L}}\;{\rm{ \times }}\;{\rm{285}}\cancel{{\rm{K}}}}}{{{\rm{298}}\cancel{{{\rm{K}}\;}}\;{\rm{1}}{\rm{.10}}\;\cancel{{{\rm{atm}}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.96}}\;{\rm{L}}$

14.

What is the density of CO2 gas at 386 K and 17.0 atm.

23.6 g/L

Solution

The correlation between density and molar mass of a gas is derived from the ideal gas law equation:

${\rm{d}}\;{\rm{ = }}\;\frac{{{\rm{PM}}}}{{{\rm{RT}}}}$

${\rm{d}}\;{\rm{ = }}\;\frac{{{\rm{17}}{\rm{.0}}\;\cancel{{{\rm{atm}}}}\;{\rm{44}}{\rm{.0}}\;{\rm{g}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}}}{{{\rm{0}}{\rm{.08206}}\;{\rm{L}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}{\rm{ 386}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{23}}{\rm{.6}}\;{\rm{g/L}}$

15.

Determine the density of ammonia gas, NH3, at 36.0 oC and 695 mmHg? Report the density in grams per liter.

0.613 g/L

Solution

The correlation between density and molar mass of a gas is derived from the ideal gas law equation:

${\rm{d}}\;{\rm{ = }}\;\frac{{{\rm{PM}}}}{{{\rm{RT}}}}$

Before plugging the numbers, convert the units to L, K, and atm to match the units of R.

T = 36.0 + 273 = 309 K

P = 695 mmHg x 1 atm/760 mmHg = 0.914 atm

${\rm{d}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.914}}\;\cancel{{{\rm{atm}}}}\;{\rm{17}}{\rm{.0}}\;{\rm{g}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}}}{{{\rm{0}}{\rm{.08206}}\;{\rm{L}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}{\rm{ 309}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.613}}\;{\rm{g/L}}$

16.

A scientist carries out an experiment to determine the molar mass of a 2.84-g sample of a colorless liquid which exerts 756 mmHg pressure when vaporized in a 260-mL flask at 142 oC. What is the molecular mass of this compound?

374 g/mol

Solution

${\rm{PV}}\;{\rm{ = }}\;{\rm{nRT}}$

${\rm{PV}}\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{M}}}{\rm{RT}}$

${\rm{M}}\;{\rm{ = }}\;\frac{{{\rm{mRT}}}}{{{\rm{PV}}}}$

Before plugging the numbers, convert the units to L, K, and atm to match the units of R.

T =142 + 273 = 415 K

P = 756 mmHg x 1 atm/760 mmHg = 0.995 atm

V = 260 mL x 1L/1000 mL = 0.260 L

${\rm{M}}\;{\rm{ = }}\;\frac{{{\rm{2}}{\rm{.84}}\;{\rm{g }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{415}}\;\cancel{{\rm{K}}}}}{{{\rm{0}}{\rm{.995}}\;\cancel{{{\rm{atm}}}}\;{\rm{0}}{\rm{.260}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{374}}\;{\rm{g/mol}}$

17.

Identify the unknown gas that weighs 17.75 grams in a 17.0 L cylinder held at 0.700 atm pressure and 250°C.
a) NO2 b) CO2 c) H2 d) SO2 e) He

M = 64.0 g/mol, SO2

Solution

To identify the unknown gas, we need to determine its molar mass.

${\rm{PV}}\;{\rm{ = }}\;{\rm{nRT}}$

${\rm{PV}}\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{M}}}{\rm{RT}}$

${\rm{M}}\;{\rm{ = }}\;\frac{{{\rm{mRT}}}}{{{\rm{PV}}}}$

Before plugging the numbers, convert the units to L, K, and atm to match the units of R.

T =250 + 273 = 523 K

${\rm{M}}\;{\rm{ = }}\;\frac{{{\rm{17}}{\rm{.75}}\;{\rm{g }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{523}}\;\cancel{{\rm{K}}}}}{{{\rm{0}}{\rm{.700}}\;\cancel{{{\rm{atm}}}}\;{\rm{17}}{\rm{.0}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{64}}{\rm{.0}}\;{\rm{g/mol}}$

18.

What is the ratio of the effusion rates of hydrogen gas (H2) and carbon dioxide (CO2) at the same pressure and temperature?

4.7

Solution

According to the Graham’s law of effusion:

$\frac{{{\rm{rate}}\;{{\rm{H}}_{\rm{2}}}}}{{{\rm{rate}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{H}}_{\rm{2}}}}}}}}$

$\frac{{{\rm{rate}}\;{{\rm{H}}_{\rm{2}}}}}{{{\rm{rate}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ = }}\;\sqrt {\frac{{{\rm{44}}{\rm{.0}}\;{\rm{g/mol}}}}{{{\rm{2}}{\rm{.0}}\;{\rm{ g/mol}}}}} \; = \;4.7$

19.

The rate of effusion of an unknown gas is 9.20 mL/min. Under identical conditions, the rate of effusion of pure nitrogen (N2) gas is 14.65 mL/min. Identify the unknown gas using the Graham’s law.
a) O2 b) C3H8 c) C4H10 d) NO2 e) Cl2

M = 71 g/mol, Cl2

Solution

According to the Graham’s law of effusion:

$\frac{{{\rm{rate}}\;{\rm{X}}}}{{{\rm{rate}}\;{{\rm{N}}_{\rm{2}}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{\rm{X}}}}}}$

Square both sides of the equation to get rid of the square root:

${\left( {\frac{{{\rm{rate}}\;{\rm{X}}}}{{{\rm{rate}}\;{{\rm{N}}_{\rm{2}}}}}} \right)^2}\;{\rm{ = }}\;\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{\rm{X}}}}}$

${{\rm{M}}_{\rm{X}}}\; = \,\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\left( {\frac{{{\rm{rate}}\;{\rm{X}}}}{{{\rm{rate}}\;{{\rm{N}}_{\rm{2}}}}}} \right)}^2}}}$

${{\rm{M}}_{\rm{X}}}\; = \,\frac{{{\rm{28}}{\rm{.0}}\;{\rm{g/mol}}}}{{{{\left( {\frac{{{\rm{9}}{\rm{.20}}\;{\rm{mL/min}}}}{{{\rm{14}}{\rm{.65}}\;{\rm{mL/min}}}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{71}}{\rm{.0}}\;{\rm{g/mol}}$

Therefore, the unknown gas is chlorine.

video

20.

A sample of krypton effuses from a container in 95 seconds. The same amount of an unknown gas requires 55 seconds. Identify the unknown gas.

Mx = 28.1 g/mol

From common gases, this can be N2 or CO.

Solution

According to the Graham’s law of effusion:

$\frac{{{\rm{rate}}\;{\rm{Kr}}}}{{{\rm{rate}}\;{\rm{X}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{\rm{X}}}}}{{{{\rm{M}}_{{\rm{Kr}}}}}}}$

Pay attention, in this problem, it is not rates that are given, it is the time each gas takes to effuse. Because, rate ∼ 1/t, we can write that:

$\frac{{{{\rm{t}}_{\rm{X}}}}}{{{{\rm{t}}_{{\rm{Kr}}}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{\rm{X}}}}}{{{{\rm{M}}_{{\rm{Kr}}}}}}}$

$\frac{{{\rm{55}}}}{{{\rm{95}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{\rm{X}}}}}{{{\rm{83}}{\rm{.8}}}}}$

${\left( {\frac{{{\rm{55}}}}{{{\rm{95}}}}} \right)^2}\;{\rm{ = }}\;\frac{{{{\rm{M}}_{\rm{X}}}}}{{{\rm{83}}{\rm{.8}}}}$

$0.335\;{\rm{ = }}\;\frac{{{{\rm{M}}_{\rm{X}}}}}{{{\rm{83}}{\rm{.8}}}}$

Mx = 28.1 g/mol

From common gases, this can be N2 or CO.

The numbers makes sense because the unknown gas is lighter and therefore, the rate of effusion is higher and it takes less time to effuse.

21.

If a sample of Br2 vapor can effuse from an opening in a heated vessel in 46 s, how long will it take the same amount of He to effuse under identical conditions?

7.3 s

Solution

According to the Graham’s law of effusion:

$\frac{{{\rm{rate}}\;{\rm{B}}{{\rm{r}}_{\rm{2}}}}}{{{\rm{rate}}\;{\rm{He}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{{\rm{He}}}}}}{{{{\rm{M}}_{{\rm{B}}{{\rm{r}}_{\rm{2}}}}}}}}$

$\frac{{{\rm{rate}}\;{\rm{B}}{{\rm{r}}_{\rm{2}}}}}{{{\rm{rate}}\;{\rm{He}}}}\;{\rm{ = }}\;\sqrt {\frac{{{\rm{4}}{\rm{.00}}}}{{{\rm{159}}{\rm{.8}}}}} \; = \;0.158$

$\frac{{\rm{1}}}{{0.158}}\;{\rm{ = }}\;6.329\;$

This is how much faster He effuses compared to Br2. Therefore, it will take that times less to effuse:

tHe = 46 s ÷ 6.329 = 7.3 s

22.

It has been demonstrated that 3.56 mL of an unknown gas effuses through a hole in the same time that 8.64 mL of argon does under the same conditions. Determine the molecular mass of the unknown gas.

236 g/mol

Solution

The volume of the gas is directly proportional to the rate of the effusion. The larger the volume, the higher the rate of the effusion. Given this information, we can determine the ratio of effusion rates of Ar and the unknown gas.

8.64 mL/3.56 mL = 2.43

So, Ar effuses 2.43 times faster than the unknown gas and we can use this to set up an equation for the Graham’s law:

$\frac{{{\rm{rate}}\;{\rm{Ar}}}}{{{\rm{rate}}\;{\rm{X}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{\rm{X}}}}}{{{{\rm{M}}_{{\rm{Ar}}}}}}} \; = \;2.43$

${\left( {2.43} \right)^2}\;{\rm{ = }}\;\frac{{{{\rm{M}}_{\rm{X}}}}}{{{{\rm{M}}_{{\rm{Ar}}}}}}\;$

Mx = 5.9049 x 39.9 = 236 g/mol

23.

What is the partial pressure of nitrogen in a mixture of N2, SO2 and CO2 that has a total pressure of 6.84 atm. The partial pressure of SO2 is 2.10 atm, and the partial pressure of COis 1.74 atm.

P(N2) = 3.00 atm

Solution

The total pressure is the sum of the partial pressures of all the gasses in the mixture:

PT = P(N2) + P(SO2) + P(CO2) = 6.84 atm

P(N2) = P– [P(SO2) + P(CO2)]

P(N2) = 6.84 atm – [2.10 atm + 1.74 atm]

P(N2) = 3.00 atm

24.

13.2 grams of COand 6.00 grams of He are mixed in a 4.00 L container at 300. K. Calculate the partial pressure of both gases and the total pressure of the mixture.

P(CO2) = 1.85 atm

P(He) = 9.23 atm

Ptotal = 11.1 atm

Solution

To determine the pressure of each gas (or the total pressure), we need to use the ideal gas law equation:

PV = nRT

${\rm{P}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{V}}}$

So, let’s calculate the moles and then the partial pressure of each gas.

n(CO2) = 13.2 g/44.0 g/mol = 0.300 mol

n (He) = 6.00 g/4.00 g/mol = 1.50 mol

${\rm{P}}\;{\rm{(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{V}}}$

${\rm{P}}\;{\rm{(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.300}}\;\cancel{{{\rm{mol}}}}{\rm{ \times }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;{\rm{atm}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{300}}{\rm{.}}\;\cancel{{\rm{K}}}}}{{{\rm{4}}{\rm{.00}}\;\cancel{{\rm{L}}}}}\; = \;{\rm{1}}{\rm{.85}}\;{\rm{atm}}$

${\rm{P}}\;{\rm{(He)}}\;{\rm{ = }}\;\frac{{{\rm{1}}{\rm{.50}}\;\cancel{{{\rm{mol}}}}{\rm{ \times }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;{\rm{atm}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{300}}{\rm{.}}\;\cancel{{\rm{K}}}}}{{{\rm{4}}{\rm{.00}}\;\cancel{{\rm{L}}}}}\; = \;{\rm{9}}{\rm{.23}}\;{\rm{atm}}$

Ptotal = 1.85 + 9.23 = 11.1 atm

25.

A 3.00-L bulb containing N2 at 1.80 atm pressure is connected to a 2.00-L bulb filled with H2 at 3.50 atm pressure. What is the final pressure of the system when the valve is opened?

2.48 atm

Solution

To solve this problem, we work with one gas at a time. For example, first, we consider that there is no H2, and only nitrogen is going to occupy both containers once the valve is open. This is a correlation between the volume and the pressure of the gas (Boyle’s law), which we can use to calculate the final pressure of the nitrogen. So, let’s write for nitrogen:

P1 = 1.80 atm

V1 = 3.00 L

V2 = 3+2 = 5.00 L

P2 = ?

P1V1 = P2V2

${{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}$

${{\rm{P}}_{\rm{2}}}\;{\rm{ = }}\;\frac{{{\rm{1}}{\rm{.80}}\;{\rm{atm 3}}{\rm{.00}}\;{\rm{L}}}}{{{\rm{5}}{\rm{.00}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.08}}\;{\rm{atm}}$

Next, we do the same for the hydrogen:

P1 = 3.50 atm

V1 = 2.00 L

V2 = 3+2 = 5.00 L

P2 = ?

${{\rm{P}}_{\rm{2}}}\;{\rm{ = }}\;\frac{{{\rm{3}}{\rm{.50}}\;{\rm{atm 2}}{\rm{.00}}\;{\rm{L}}}}{{{\rm{5}}{\rm{.00}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.4}}\;{\rm{atm}}$

The total pressure will be the sum of these partial pressures:

PTotal = 1.08 + 1.40 = 2.48 atm

video

26.

Assuming ideal gas behavior, calculate the total pressure (in atm) of a mixture of 0.0260 mol of nitrogen, N2, and 0.0170 mol of argon, Ar, in a 3.50-L flask at 20 oC.

0.295 atm

Solution

To determine the total pressure of the gases, we need to use the ideal gas law equation:

PV = nRT

${\rm{P}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{V}}}$

So, let’s calculate the total moles of the gases.

n(total) = n(N2) + n (Ar) = 0.0260 + 0.0170 = 0.0430 mol

${\rm{P}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{V}}}$

${\rm{P}}\;{\rm{(total)}}\;{\rm{ = }}\;\frac{{0.0430\;\cancel{{{\rm{mol}}}}{\rm{ \times }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;{\rm{atm}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{293}}\;\cancel{{\rm{K}}}}}{{{\rm{3}}{\rm{.50}}\;\cancel{{\rm{L}}}}}\; = \;{\rm{0}}{\rm{.295}}\;{\rm{atm}}$

27.

A mixture of gases contains CH4, N2, and H2 and exerts a total pressure of 2.65 atm. The mixture contains 0.456 mol of CH4, 0.540 mol of N2 and 0.730 mol of H2. What is the partial pressure of hydrogen in atmospheres?

1.12 atm

Solution

The relationship between partial pressure and the total pressure can be given using the mole fraction of a given gas. For hydrogen, for example, it is:

P(H2) = χ(H2) x Ptotal

So, to calculate the partial pressure of hydrogen, we need to find its mole fraction in the mixture:

${\rm{\chi }}\left( {{{\rm{H}}_{\rm{2}}}} \right)\; = \;\frac{{{\rm{n(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}}}{{{\rm{n(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\; + \;{\rm{n(C}}{{\rm{H}}_{\rm{4}}}{\rm{)}}\; + \;{\rm{n(}}{{\rm{N}}_{\rm{2}}}{\rm{)}}}}$

${\rm{\chi }}\left( {{{\rm{H}}_{\rm{2}}}} \right)\; = \;\frac{{{\rm{0}}{\rm{.730}}}}{{{\rm{0}}{\rm{.730}}\; + \;0.456\; + \;{\rm{0}}{\rm{.540}}}}\; = \;0.423$

P(H2) = 0.423 x 2.65 atm = 1.12 atm

28.

The partial pressures of CH4, C3H8, and C4H10 in a gas mixture are 270 torr, 1016 torr, and 1142 torr, respectively. What is the mole fraction of butane (C4H10)?

0.470

Solution

The relationship between partial pressure and the total pressure can be given using the mole fraction of a given gas. For butane, for example, it is:

P(C4H10) = χ(C4H10) x Ptotal

${\rm{\chi }}\left( {{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}} \right)\;{\rm{ = }}\;\frac{{{\rm{P}}\left( {{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}} \right)}}{{{{\rm{P}}_{{\rm{total}}}}}}$

${\rm{\chi }}\left( {{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}} \right)\;{\rm{ = }}\;\frac{{{\rm{1142}}}}{{{\rm{270}}\;{\rm{ + }}\;{\rm{1016}}\;{\rm{ + }}\;{\rm{1142}}}}\; = \;0.470$

29.

In a mixture of two gases, the partial pressure of CO2(g) is 0.145 atm and that of O2(g) is 0.370 atm.

a) What is the mole fraction of each gas in the mixture?

b) Calculate the total number of moles of gas in the mixture if the mixture occupies a volume of 12.5 L at 45.0 °C.

c) Calculate the number of grams of each gas in the mixture.

a) χ(CO2) = 0.282 , χ(O2) = 0.718

b) 0.247 mol

c) m(CO2) = 3.06 g, m(O2) = 5.68 g

Solution

a) The relationship between partial pressure and the total pressure can be given using the mole fraction of a given gas.

${\rm{\chi }}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{{\rm{P}}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)}}{{{{\rm{P}}_{{\rm{total}}}}}}$

${\rm{\chi }}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.145}}\;{\rm{atm}}}}{{{\rm{0}}{\rm{.145}}\;{\rm{atm}}\;{\rm{ + }}\;{\rm{0}}{\rm{.370 atm}}}}\; = \;0.282$

The mole fraction of Othen is 1-0.282 = 0.718

Let’s check this with a calculation:

${\rm{\chi }}\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.370}}\;{\rm{atm}}}}{{{\rm{0}}{\rm{.145}}\;{\rm{atm}}\;{\rm{ + }}\;{\rm{0}}{\rm{.370 atm}}}}\; = \;0.718$

b) To find the total number of moles, we use the ideal gas law equation.

PV = nRT

${\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{PV}}}}{{{\rm{RT}}}}$

Don’t forget to convert oC to K:

T = 45.0+273 = 318 K

${\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.515}}\;\cancel{{{\rm{atm}}}}\;{\rm{12}}{\rm{.5}}\,\cancel{{\rm{L}}}}}{{{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{318}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.247}}\;{\rm{mol}}$

Notice that we have used the total pressure because we are looking at the total moles of gases.

c) We can calculate the mass of each gas by using its moles. These, in turn, can be calculated using the mole fraction of each gas.

${\rm{\chi }}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\; = \;\frac{{{\rm{n(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}}}{{{\rm{n(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\; + \;{\rm{n(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}}}$

${\rm{0}}{\rm{.282}}\;{\rm{ = }}\;\frac{{{\rm{n(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}}}{{{\rm{0}}{\rm{.247}}\;{\rm{mol}}}}$

n(CO2) = 0.282 x 0.247 mol = 0.0697 mol

m(CO2) = 0.0697 mol x 44.0 g/mol = 3.06 g

${\rm{\chi }}\left( {{{\rm{O}}_{\rm{2}}}} \right)\; = \;\frac{{{\rm{n(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}}}{{{\rm{n(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\; + \;{\rm{n(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}}}$

${\rm{0}}{\rm{.718}}\;{\rm{ = }}\;\frac{{{\rm{n(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}}}{{{\rm{0}}{\rm{.247}}\;{\rm{mol}}}}$

n(O2) = 0.718 x 0.247 mol = 0.177 mol

m(O2) = 0.177 mol x 32.0 g/mol = 5.68 g

30.

Oxygen can be prepared in the laboratory by decomposing potassium chlorate, KClO3 according to the following reaction:

2KClO3(s) → 2KCl(s) + 3O2(g).

How many liters of oxygen can be produced from 4.80 g KClO3, if the reaction is carried out at standard conditions?

1.32 L

Solution

To determine the volume of oxygen produced in this reaction, we need to first find how many moles of it are produced from 4.80 g KClO3.

This is done based on the stoichiometry of the reaction. So, let’s first see how many moles of KClOwe have:

n(KClO3) = 4.80 g/122.6 g/mol = 0.0392 mol

Next, we find the moles of oxygen based on the stoichiometric ratio:

${\rm{n(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0392}}\;{\rm{mol}}\;{\rm{KCl}}{{\rm{O}}_{\rm{3}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{2}}\;{\rm{mol}}\;{\rm{KCl}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0587}}\;{\rm{mol}}$

After this, we use the ideal gas law equation to determine the volume of the oxygen:

PV = nRT

${\rm{V}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{P}}}$

${\rm{V}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.0587}}\;\cancel{{{\rm{mol}}}}\;{\rm{0}}{\rm{.08206}}\;{\rm{L}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}\;{\rm{273}}\;\cancel{{\rm{K}}}}}{{{\rm{1}}{\rm{.00}}\;\cancel{{{\rm{atm}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.32}}\;{\rm{L}}$

video

31.

Sodium azide is used in automobile airbags as a source of nitrogen gas to rapidly inflate the bags upon a hit. If the volume of the airbag is 7.95 L, what mass of NaN3 is required to produce enough nitrogen to fill it at 23.0 °C and 1.20 atm?

2NaN3(s)  →  2Na(s)  +  3N2(g)

17.0 g

Solution

The plan for solving this problem is to find the moles of nitrogen, use that to determine the moles of sodium azide based on the stoichiometric ratio, and finally convert the moles of NaN3 to the mass that is needed for producing the given amount of nitrogen gas.

So, first thing, determine the moles of nitrogen using the ideal gas law equation:

PV = nRT

${\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{PV}}}}{{{\rm{RT}}}}$

${\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{1}}{\rm{.20 }}\cancel{{{\rm{atm}}}}{\rm{ 7}}{\rm{.95 }}\cancel{{\rm{L}}}}}{{{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}\;}}{\rm{296}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.393}}\;{\rm{mol}}$

Next, determine the moles of NaN3:

${\rm{n(Na}}{{\rm{N}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.393}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{Na}}{{\rm{N}}_{\rm{3}}}}}{{{\rm{3}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.262}}\;{\rm{mol}}$

The last step is simply converting the moles to the mass of NaN3:

m(NaN3) = 0.262 mol x 65.0 g/mol = 17.0 g

video

32.

Calcium hydride (CaH2) can be used as a drying agent and a source of Hydrogen, because of its high reactivity with water.

CaH2(s) + 2 H2O(l)  →  Ca(OH)2(aq) + 2H2(g)

How many grams of CaH2 are needed to produce 76.8 L of H2 gas at a pressure of 0.750 atm and a temperature of 25 °C?

49.6 g

Solution

The plan for solving this problem is to find the moles of hydrogen, use that to determine the moles of calcium hydride based on the stoichiometric ratio, and finally convert the moles of CaH2 to the mass that is needed for producing the given amount of hydrogen gas.

So, first thing, determine the moles of nitrogen using the ideal gas law equation:

PV = nRT

${\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{PV}}}}{{{\rm{RT}}}}$

${\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.750 }}\cancel{{{\rm{atm}}}}{\rm{ 76}}{\rm{.8 }}\cancel{{\rm{L}}}}}{{{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}\;}}{\rm{298}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.36}}\;{\rm{mol}}$

Next, determine the moles of CaH2:

${\rm{n(Ca}}{{\rm{H}}_2}{\rm{)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.36}}\;{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{Ca}}{{\rm{H}}_2}}}{{{\rm{2}}\;{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.18}}\;{\rm{mol}}$

And now convert this to the mass of CaH2:

m(CaH2) = 1.18 mol x 42.1 g/mol = 49.6 g

33.

Propane (C3H8) is used as a fuel in gas barbecue grills. How many liters of oxygen, taken at STP, are needed for the full combusting of 26.4 g of propane?

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

67.0 L

Solution

The plan is to find the moles of oxygen based on the stoichiometric ratio with propane and then determine the volume using the ideal ga law equation.

n(C3H8) = 26.4 g /44.1 g/mol = 0.599 mol

The moles of oxygen are 5 times more (1:5 ratio).

n(O2) = 5 x 0.599 mol = 2.99 mol

Of course, this can also be done by mole ration conversion:

${\rm{n(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.599}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}\;{\rm{ \times }}\;\frac{{{\rm{5}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{1}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.99}}\;{\rm{mol}}$

And now, we can convert this to volume using the ideal gas law equation:

${\rm{V}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{P}}}$

${\rm{V}}\;{\rm{ = }}\;\frac{{{\rm{2}}{\rm{.99}}\;\cancel{{{\rm{mol}}}}\;{\rm{0}}{\rm{.08206}}\;{\rm{L}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}\;{\rm{273}}\;\cancel{{\rm{K}}}}}{{{\rm{1}}{\rm{.00}}\;\cancel{{{\rm{atm}}}}}}\;{\rm{ = }}\;{\rm{67}}{\rm{.0}}\;{\rm{L}}$

34.

Lithium hydroxide (LiOH) is used in space shuttles to absorb the carbon dioxide exhaled by astronauts according to the following equation:

2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)

How many liters of carbon dioxide gas at 23.0 oC and 732 mmHg could be absorbed by 284 g of lithium hydroxide?

149 L

Solution

The first thing is to find the moles of LiOH and based on the stoichiometric ratio, determine the moles of CO2.

n(LiOH) = 284 g x 24.0 g/mol = 11.8 mol

${\rm{n(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{11}}{\rm{.8}}\;{\rm{mol}}\;{\rm{LiOH}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{2}}\;{\rm{mol}}\;{\rm{LiOH}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.90}}\;{\rm{mol}}$

${\rm{V}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{P}}}$

Remember to convert the temperature to K:

T = 23.0 + 273 = 296 K

The pressure must be in atm since R contains atm.

P = 732 mmHg x 1 atm/760 mmHg = 0.963 atm

${\rm{V}}\;{\rm{ = }}\;\frac{{{\rm{5}}{\rm{.90}}\;\cancel{{{\rm{mol}}}}\;{\rm{0}}{\rm{.08206}}\;{\rm{L}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}\;{\rm{296}}\;\cancel{{\rm{K}}}}}{{{\rm{0}}{\rm{.963}}\;\cancel{{{\rm{atm}}}}}}\;{\rm{ = }}\;{\rm{149}}\;{\rm{L}}$

35.

In a particular reaction, at 25 °C, 16.4 L of carbon monoxide at a 950.0 torr is mixed with 11.2 g of iron (III) oxide, and 5.68 g iron is obtained. What is the percent yield of the reaction?

Fe2O3 + 3CO(g) → 2Fe + 3CO2(g)

72.9%

Solution

If the problem involves stoichiometry and percent yield, you would be on the right track determining the moles of the reactants and products first.

The moles of CO can be calculated by using the ideal gas law equation.

PV = nRT

${\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{PV}}}}{{{\rm{RT}}}}$

Before plugging the numbers, we do the necessary conversions of the temperature and pressure.

T = 25.0 + 273 = 298 K

P = 950.0 torr x 1 atm/ 760 torr = 1.25 atm

${\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{1}}{\rm{.25 }}\cancel{{{\rm{atm}}}}{\rm{ 16}}{\rm{.4 }}\cancel{{\rm{L}}}}}{{{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}\;}}{\rm{298}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.838}}\;{\rm{mol}}$

The moles of Fe2O3 are calculated from its mass and molar mass:

n (Fe2O3) = 11.2 g/159.7 g/mol = 0.0701 mol

The moles of Fe produced in the reaction is also calculated from its mass and molar mass:

n (Fe) = 5.68 g/55.8 g/mol = 0.102 mol

This is the actual yield of the reaction, and to find the percent yield, we need to determine how much iron could have been produced based on the limiting reactant. Remember, the limiting reactant is the one that gives less product. So, to find the LR, we determine whether the Fe2O3 or CO could produce less product based on their moles and the reaction stoichiometry.

${\rm{n}}\left( {{\rm{Fe}}\;{\rm{ from }}\;{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}} \right)\;\;{\rm{ = }}\;{\rm{0}}{\rm{.0701}}\;{\rm{mol}}\;{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{Fe}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.140}}\;{\rm{mol}}$

${\rm{n}}\left( {{\rm{Fe}}\;{\rm{ from }}\;{\rm{CO}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.838}}\;{\rm{mol}}\;{\rm{CO}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{Fe}}}}{{{\rm{3}}\;{\rm{mol}}\;{\rm{CO}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.559}}\;{\rm{mol}}$

Therefore, Fe2O3 is the limiting reactant and the theoretical yield of the reaction is 0.140 mol Fe. The percent yiled of the reaction is:

${\rm{\% }}\;{\rm{yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\; \times \;100\% \;$

${\rm{\% }}\;{\rm{yield}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.102}}}}{{{\rm{0}}{\rm{.140}}}}\; \times \;100\% \; = \;72.9\%$