## General Chemistry

These calculations are based on the same principles we learn in stoichiometric calculations. The only difference for most problems is going to be the fact that the quantities are given in molarities rather than in moles. Remember that you need to always calculate the moles for solving reaction stoichiometry problems.

For example,

How many ml of 1.80 M MgBr2 is necessary to precipitate all the silver ions in 600.0 mL of 0.80 M AgNO3 solution?

Write a balanced chemical equation for this reaction and follow the plan for the calculation which is to determine the moles, do the stoichiometric calculations, and convert to volume or any other quantity asked in the problem:

2AgNO3(aq) + MgBr2(aq) → 2AgBr(s) + Mg(NO3)2(aq)

The plan is to calculate the moles of AgNO3, then determine the moles of MgBr2 based on the stoichiometric ratio, and finally convert the moles of MgBr2 to mL using the formula for molarity.

To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. So, convert the mL to L first:

${\rm{V}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{600}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.6000}}\;{\rm{L}}$

Next, we rearrange the formula for molarity to calculte the moles of AgNO3:

${\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}$

n = MV

${\rm{n}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.80}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.6000}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.48}}\;{\rm{mol}}$

Now, we use this to determine the moles of MgBr2 based on the stoichiometric ratio:

${\rm{n}}\left( {{\rm{MgB}}{{\rm{r}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.48}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MgB}}{{\rm{r}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.24}}\;{\rm{mol}}$

An finally, convert the moles of MgBrto the volume of its solution using the molarity formula:

${\rm{V}}\left( {{\rm{MgB}}{{\rm{r}}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.24}}\;{\rm{mol}}}}{{{\rm{1}}{\rm{.80}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.13}}\;{\rm{L}}$

To convert this to mL, we multiply it by 1000, so it is 130 mL.

Check Also

#### Practice

1.

How many ml of 1.80 M MgBr2 is necessary to precipitate all the silver ions in 600.0 mL of 0.80 M AgNO3 solution?

130 mL

Solution

The first step is to write a balanced chemical equation for this reaction:

2AgNO3(aq) + MgBr2(aq) → 2AgBr(s) + Mg(NO3)2(aq)

The plan is to calculate the moles of AgNO3, then determine the moles of MgBr2 based on the stoichiometric ratio, and finally convert the moles of MgBr2 to mL using the formula for molarity.

To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. . So, convert mL to L first:

${\rm{V}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{600}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.6000}}\;{\rm{L}}$

${\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}$

n = MV

${\rm{n}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.80}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.6000}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.48}}\;{\rm{mol}}$

Now, we use this to determine the moles of MgBr2 based on the stoichiometric ratio:

${\rm{n}}\left( {{\rm{MgB}}{{\rm{r}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.48}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MgB}}{{\rm{r}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.24}}\;{\rm{mol}}$

${\rm{V}}\left( {{\rm{MgB}}{{\rm{r}}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.24}}\;{\rm{mol}}}}{{{\rm{1}}{\rm{.80}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.13}}\;{\rm{L}}$

To convert this to mL, we multiply it by 1000, so it is 130 mL.

2.

How many ml of 2.70 M K2SO4 is required to precipitate all the barium ions in 500. mL of 0.650 M BaCl2 solution?

120 mL

Solution

The first step is to write a balanced chemical equation for this reaction:

K2SO4 (aq) + BaCl2 (aq) → BaSO4 (s) + 2KCl(aq)

The plan is to calculate the moles of BaCl2, then determine the moles of K2SO4 based on the stoichiometric ratio, and finally convert the moles of K2SO4 to mL using the formula for molarity.

To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. So, convert mL to L first:

${\rm{V}}\left( {{\rm{BaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;5{\rm{00}}{\rm{.}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.500}}\;{\rm{L}}$

${\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}$

n = MV

${\rm{n}}\left( {{\rm{BaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.650}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.500}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.325}}\;{\rm{mol}}$

Now, we use this to determine the moles of K2SO4 based on the stoichiometric ratio:

${\rm{n}}\left( {{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.325}}\;\cancel{{{\rm{mol}}\;{\rm{BaC}}{{\rm{l}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{BaC}}{{\rm{l}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.325}}\;{\rm{mol}}$

Next, use the moles in the formula for molarity and determine the volume of the solution in L:

${\rm{V}}\left( {{\rm{MgB}}{{\rm{r}}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.325}}\;{\rm{mol}}}}{{{\rm{2}}{\rm{.70}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.120}}\;{\rm{L}}$

To convert this to mL, we multiply it by 1000, so it is 120 mL.

3.

27.0 mL of a NaOH solution was needed to neutralize 0.326 g of KHP (KHC8H4O4). What is the molarity of the NaOH solution?

0.0593 mol/L

Solution

First, let’s write a balanced chemical equation for this reaction:

NaOH + KHC8H4O4 → NaKC8H4O4 + H2O

The plan here would be calculating the moles of KHP, using it to determine the moles of NaOH based on the stoichiometric ration, and use that to calculate the volume of NaOH solution:

${\rm{n}}\left( {{\rm{KHP}}} \right)\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.326}}\;{\rm{g}}}}{{{\rm{204}}{\rm{.2}}\;{\rm{g/mol}}}}{\rm{ = }}\;{\rm{0}}{\rm{.00160}}\;{\rm{mol}}$

Because the ratio of NaOH and KHP is 1:1, the moles are going to be equal and therefore, there are 0.00160 mol NaOH reacting with the KHP.

Now, we can use the moles and the volume of the solution to determine the molarity of NaOH. Don’t forget to convert mL to L when working with molarity:

${\rm{V}}\left( {{\rm{NaOH}}} \right)\;{\rm{ = }}\;27.0\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0270}}\;{\rm{L}}$

${\rm{M}}\left( {{\rm{NaOH}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.00160}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.0270}}\;{\rm{L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0593}}\;{\rm{mol/L}}$

4.

What volume of 0.150 M calcium hydroxide is required to neutralize 250.0 mL of 0.0350 M sulfuric acid?

0.0583 L

Solution

First, let’s write a balanced chemical equation for this reaction:

Ca(OH)2(s) + H2SO4(aq) → CaSO4(aq) + 2H2O(l)

The linkage between the two reactants/solutions is still going to be the mole conversion based on the stoichiometric ratio. So, to plan is as follows:

To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. So, convert mL to L first:

${\rm{V}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{250}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.2500}}\;{\rm{L}}$

${\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}$

n = MV

${\rm{n}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0350}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.2500}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00875}}\;{\rm{mol}}$

Now, we use this to determine the moles of Ca(OH)2 based on the stoichiometric ratio:

${\rm{n}}\left( {{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.00875}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00875}}\;{\rm{mol}}$

Next, use the moles in the formula for molarity and determine the volume of the solution in L:

${\rm{V}}\left( {{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.00875}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.150}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0583}}\;{\rm{L}}$

5.

How many mL of 0.250 M HCl will neutralize 60.00 mL of 0.060 M Ba(OH)2?

28.8 mL

Solution

First, let’s write a balanced chemical equation for this reaction:

Ba(OH)2(s) + 2HCl(aq) → BaCl2(aq) + 2H2O(l)

The linkage between the two reactants/solutions is still going to be the mole conversion based on the stoichiometric ratio. So, the plan is as follows:

To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. So, convert mL to L first:

${\rm{V}}\left( {{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{60}}{\rm{.00}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.06000}}\;{\rm{L}}$

${\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}$

n = MV

${\rm{n}}\left( {{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.060}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.06000}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0036}}\;{\rm{mol}}$

Now, we use this to determine the moles of HCl based on the stoichiometric ratio:

${\rm{n}}\left( {{\rm{HCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0036}}\;\cancel{{{\rm{mol}}\;{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{HCl}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0072}}\;{\rm{mol}}$

Next, use the moles in the formula for molarity and determine the volume of the solution in L and convert it to mL:

${\rm{V}}\left( {{\rm{HCl}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.0072}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.250}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0288}}\;{\rm{L}}$

To get the mL, we just need to multiply by 1000, or use the molar conversion:

${\rm{V}}\left( {{\rm{HCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0288}}\;{\rm{L}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{mL}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{28}}{\rm{.8}}\;{\rm{mL}}$

6.

What is the mass in grams of AgCl precipitate if 20.0 mL of 0.150 M AgNO3 is added to 45.0 mL of 0.250 M MgCl2?

0.430 g

Solution

The first step is to write a balanced chemical equation for this reaction:

2AgNO3(aq) + MgCl2(aq) → 2AgCl(s) + Mg(NO3)2(aq)

Next, calculate the moles and determine the limiting reactant. For this, we need to use the formula for molarity:

n = MV

Before entering the numbers, remember to convert the volume to liters because the unit for molarity is mol/L.

${\rm{V}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{20}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0200}}\;{\rm{L}}$

${\rm{V}}\left( {{\rm{MgC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{45}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0450}}\;{\rm{L}}$

And now, we can use the volumes and concentrations to calculate the moles:

${\rm{n}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.150}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0200}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00300}}\;{\rm{mol}}$

${\rm{n}}\left( {{\rm{MgC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.250}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0450}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0113}}\;{\rm{mol}}$

Remember, limiting reactant is the one that gives less product. We can do the calculations base don any product, but since we need to determine the mass of AgCl produced, let’s do the calculations based on it:

${\rm{n}}\left( {{\rm{AgCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.00300}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{AgCl}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00300}}\;{\rm{mol}}$

${\rm{n}}\left( {{\rm{AgCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0113 }}\cancel{{{\rm{mol}}\;{\rm{MgC}}{{\rm{l}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{AgCl}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{MgC}}{{\rm{l}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0226}}\;{\rm{mol}}$

AgNO3 produces less product and therefore, it is the limiting reactant, and we must do our calculations based on it.

The last step is to convert the moles of AgCl to mass:

${\rm{m}}\left( {{\rm{AgCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.00300}}\;\cancel{{{\rm{mol}}}}\;{\rm{ \times }}\;{\rm{143}}{\rm{.3}}\;\frac{{\;{\rm{g}}}}{{\cancel{{{\rm{mol}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.430}}\;{\rm{g}}$

7.

How many mL of 0.600 M K2SO4 was added to 400.0 mL of 0.500 M BaCl2 solution if 31.2 g of precipitate was collected?

223 mL

Solution

First, write a balanced chemical equation for this reaction:

K2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2KCl(aq)

The amount of product formed is given, so we know the actual yield of the reaction. This means our calculations for the amounts of reactants that reacted must be done based on the moles of the product. The precipitate is BaSO4, and to determine its moles, we use the formula relating the mass, molar mass, and the moles:

${\rm{n(BaS}}{{\rm{O}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;\frac{{{\rm{31}}{\rm{.2 g}}}}{{{\rm{233}}{\rm{.4}}\;{\rm{g/mol}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.134}}\;{\rm{mol}}\;$

This already is enough to calculate the volume of the K2SO4 solution because we can determine the moles of K2SO4 based on the stoichiometric ratio. Every one mole of K2SO4 produces one mole of BaSO4 and therefore, 0.134 mol K2SO4 had reacted.

Next, use the moles in the formula for molarity and determine the volume of the solution in L and convert it to mL:

${\rm{V}}\left( {{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.134}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.600}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.223}}\;{\rm{L}}$

To get the mL, we just need to multiply by 1000, or use the molar conversion:

${\rm{V}}\left( {{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.223}}\;{\rm{L}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{mL}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{223}}\;{\rm{mL}}$

To confirm that BaCl2 was un excess, we determine its moles from the concentration and volume which is 0.4000 L (400.0 mL):

${\rm{n(BaC}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.500}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.4000}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.200}}\;{\rm{mol}}$

Now, 0.200 mol of BaCl2 would have produced as much of BaSO4 since they have a 1:1 molar ratio. This amount is greater than what was formed from 0.134 mol K2SO4 which proves that BaCl2 was present in excess and K2SO4 was the limiting reactant.

8.

When a 42.0-mL sample of a 0.85 M sodium chloride solution is mixed with 17.0 mL of a 0.650 M lead(II) nitrate solution, 2.45 g of a precipitate is formed:

2NaCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2NaNO3(aq)

What are the theoretical and the percent yields of the reaction?

Theoretical yield = 3.09 g

% yield = 79.8%

Solution

The quantities of both reactants are given, so we need to determine the limiting reactant. Remember, limiting reactant is the one that gives less product. We can do the calculations based on any product, but since we are working in PbCl2, let’s do the calculations based on it.

The first step in doing these calculations, if to find the moles of both reactants for which we use the formula for molarity. As always, convert the mL to L first:

${\rm{V}}\left( {{\rm{NaCl}}} \right)\;{\rm{ = }}\;{\rm{42}}{\rm{.0}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;{\rm{mL}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.042}}\;{\rm{L}}$

${\rm{V}}\left( {{\rm{Pb}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{17}}{\rm{.0}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;{\rm{mL}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.017}}\;{\rm{L}}$

${\rm{n(NaCl)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.85}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0420}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0357}}\;{\rm{mol}}$

${\rm{n(Pb}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.650}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0170}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0111}}\;{\rm{mol}}$

Now, let’s see which of these produces less PbCl2:

${\rm{n}}\left( {{\rm{PbC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0357}}\;\cancel{{{\rm{mol}}\;{\rm{NaCl}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PbC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{NaCl}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0179}}\;{\rm{mol}}$

${\rm{n}}\left( {{\rm{PbC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0111}}\;\cancel{{{\rm{mol}}\;{\rm{Pb}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PbC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Pb}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0111}}\;{\rm{mol}}$

Pb(NO3)2 is the limiting reactant because it gives less product.

The theoretical yield of the reaction, therefore, is 0.0111 mol which is:

${\rm{m}}\left( {{\rm{PbC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0111}}\;\cancel{{{\rm{mol}}}}\;{\rm{ \times }}\;\frac{{{\rm{278}}{\rm{.1}}\;{\rm{g}}}}{{\cancel{{{\rm{mol}}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.09}}\;{\rm{g}}$

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

${\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{2}}{\rm{.45}}\;{\rm{g}}}}{{{\rm{3}}{\rm{.09}}\;{\rm{g}}}}\; \times \;100\% \; = \;79.3\;\%$

9.

A 4.598-g sample of acetylsalicylic acid (the active ingredient in aspirin) was dissolved in water. It took 35.6 mL of a 0.717 M NaOH solution to neutralize the acid. Determine the molar mass of the acid considering that it is a monoprotic acid.

180.3 g/mol

Solution

The plan here would be to find the moles of acetylsalicylic acid based on the amount of NaOH, and use the moles together with the given mass (4.598 g) to determine the molar mass:

We don’t know the formula of acetylsalicylic acid, but we know that it is a monoprotic acid and therefore, the stoichiometric ratio with NaOH is going to be 1:1. To find the moles of NaOH, we are going to use the formula for molarity so, first, express the volume in L:

${\rm{V}}\left( {{\rm{NaOH}}} \right)\;{\rm{ = }}\;{\rm{35}}{\rm{.6}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;{\rm{mL}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0356}}\;{\rm{L}}$

${\rm{n(NaOH)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.717}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0356}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0255}}\;{\rm{mol}}$

Because the moles of the acid and NaOH are equal, 0.0255 mol of acid must’ve reacted. And at this point we have the moles and the mass of the acid. Use these to determine the molar mass:

${\rm{M}}\left( {{\rm{acid}}} \right)\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{n}}}\;{\rm{ = }}\;\frac{{{\rm{4}}{\rm{.598}}\;{\rm{g}}}}{{{\rm{0}}{\rm{.0255}}\;{\rm{mol}}}}{\rm{ = }}\;{\rm{180}}{\rm{.3}}\;g/{\rm{mol}}$

10.

A student prepared a solution by dissolving 2.40-g mixture of calcium nitrate and calcium chloride. To this, a solution of silver nitrate was added dropwise until the mass of the precipitate stayed constant at 0.584 g. Determine the mass percent of calcium nitrate in the mixture.

61.4 %

Solution

The only precipitate that can be formed is AgCl when AgNO3 reacts with the CaCl2. So, if we write a balanced chemical equation, we can determine the moles and the mass of the CaCl2 in the mixture. Once we know the mass of CaCl2, we subtract it from the total mass of the mixture and get the mass of Ca(NO3)2. This, in turn, is divided by the mass of the mixture to get the percentage of Ca(NO3)2.

Let’s write the chemical equation:

2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq)

Now, calculate the moles of AgCl and determine the moles of CaCl2 based on the stoichiometric ratio:

${\rm{n}}\left( {{\rm{AgCl}}} \right)\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{2}}{\rm{.40}}\;{\rm{g}}}}{{{\rm{143}}{\rm{.3}}\;{\rm{g/mol}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0167}}\;{\rm{mol}}$

${\rm{n}}\left( {{\rm{CaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0167}}\;\cancel{{{\rm{mol}}\;{\rm{AgCl}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{CaC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{AgCl}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00835}}\;{\rm{mol}}$

The mass can be calculated using the moles and the molar mass:

${\rm{m}}\left( {{\rm{CaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.00835}}\;\cancel{{{\rm{mol}}}}\;{\rm{ \times }}\;\frac{{{\rm{111}}\;{\rm{g}}}}{{\cancel{{{\rm{mol}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.927}}\;{\rm{g}}$

m(Ca(NO3)2)= 2.40 – 0.927 = 1.473 g

${\rm{\% }}\;{\rm{Ca}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}\;{\rm{ = }}\;\frac{{{\rm{1}}{\rm{.473}}\;{\rm{g}}}}{{{\rm{2}}{\rm{.40}}\;{\rm{g}}}}\;{\rm{ \times }}\;{\rm{100\% }}\,{\rm{ = }}\;{\rm{61}}{\rm{.4\% }}$

11.

Determine the products and calculate the mass of the precipitate that is formed after 177.2 mL of a 1.25 M Na2CrO4(aq) solution is mixed with 250 mL of a 1.25 M AgNO3(aq) solution.

AgNO3(aq) + Na2CrO4(aq) →