## General Chemistry

#### Practice

1.

Provide a balanced chemical equation for the dissociation of each compound:

a) Na3PO4,  b) K2CrO4,  c) MgCl2,  d) Al(NO3)3,  e) HNO3,  f) K2SO4,  g) Ca(NO2)2,  h) Mg(C2H3O2)2,  i) Na2CO3,  j) (NH4)2SO4,  k) HClO4,  l) NaHSO3,

a)

The key part in writing dissociation equations is recognizing the ions. You must know the formula and the charge for each ion in the salt in order to avoid having wrong formulas.

Na3PO4(aq) → 3Na+(aq) + PO43-(aq)

b)

K2CrO(aq) → 2K+(aq) + CrO42-(aq)

c)

MgCl2(aq) → Mg2+(aq) + 2Cl(aq)

d)

Al(NO3)3(aq) → Al3+(aq)  + 3NO3(aq)

e)

HNO3(aq) → H+(aq) +NO3(aq)

f)

K2SO4(aq) → 2K+(aq) + SO42-(aq)

g)

Ca(NO2)2(aq) → Ca2+(aq) + 2NO2(aq)

h)

Mg(C2H3O2)2(aq) → Mg2+(aq) + 2C2H3O2(aq)

i)

Na2CO3(aq) → 2Na+(aq) + CO32-(aq)

j)

(NH4)2SO4(aq) → 2NH4+(aq) + SO42-(aq)

k)

HClO4(aq) → H+(aq) + ClO4(aq)

l)

NaHSO3(aq) → Na+(aq) + HSO3(aq)

2.

Write balanced complete ionic and net ionic equations for each reaction:

a) K2SO4(aq) + CaCl2(aq) →CaSO4(s) + 2KCl(aq)

b) Na2S(aq) + 2HBr(aq) → 2NaBr(aq) + H2S(g)

c) NH4Cl(aq) + KOH(aq) → KaCl(aq) + H2O(l) + NH3(g)

d) Na2CO3(s) + H2SO4(aq) → Na2SO4(aq) + CO2(g)+ H2O(l)

e) AgNO3(aq) + NaI(aq) → Agl(s) + NaNO3(aq)

f) HBr(aq) + KOH(aq) → KBr(aq) + H2O(l)

g) HC2H3O2(aq) + Na2CO3(aq) → NaC2H3O2(aq) + CO2(g) + H2O(l)

h) LiOH(aq) + HC2H3O2(aq) → LiC2H3O2(aq) + H2O(l)

i) HCl(aq) + NH3(aq) → NH4Cl(aq)

j) 2K3PO4(aq) + 3BaCl2(aq) → Ba3(PO4)2(s) + 6KCl(aq)

a)

What makes this exercise easier is that the states of the reactants and products are given. You need to remember that only the ones with the notation “aq” are strong electrolytes and can be dissociated.

Ionic – dissociate the electrolytes:

2K+(aq) + SO42-(aq) + Ca2+(aq) + 2Cl(aq) → CaSO4(s) + 2K+ + 2Cl(aq)
Spectator ions

Net ionic – omit spectator ions:

Ca2+(aq) + SO42-(aq) → CaSO4(s)

b)

Ionic – dissociate the electrolytes:

2Na+(aq) + S2-(aq) + 2H+ + Br(aq)2Na+(aq) + Br(aq) + H2S(g)

Net ionic – omit spectator ions:

2H+ + S2-(aq) → H2S(g)

c)

Ionic – dissociate the electrolytes:

NH4+(aq) + Cl(aq) + K+(aq) + OH(aq) → Ka+(aq) + Cl(aq) + H2O(l) + NH3(g)

Net ionic – omit spectator ions:

NH4+(aq) + OH(aq) → H2O(l) + NH3(g)

d)

Ionic – dissociate the electrolytes:

2Na+ + CO32-(s) + 2H+ + SO42-(aq)2Na+ + SO42-(aq) + CO2(g)+ H2O(l)

Net ionic – omit spectator ions:

2H+ + CO32-(s) → CO2(g) + H2O(l)

e)

Ionic – dissociate the electrolytes:

Ag+(aq) + NO3(aq) + Na+(aq) + I(aq) → Agl(s) + Na+(aq) + NO3(aq)

Net ionic – omit spectator ions:

Ag+(aq) + I(aq) → Agl(s)

f)

Ionic – dissociate the electrolytes:

H+(aq) + Br(aq) + K+(aq) + OH(aq)K+(aq) + Br(aq) + H2O(l)

Net ionic – omit spectator ions:

H+(aq) + OH(aq) → H2O(l)

g)

Ionic – dissociate the electrolytes:

2H+(aq) + C2H3O2(aq) + 2Na+(aq) + CO32(aq) → 2Na+(aq) + C2H3O2(aq) + CO2(g) + H2O(l)

Net ionic – omit spectator ions:

2H+(aq) + CO32(aq) → CO2(g) + H2O(l)

h)

Ionic – dissociate the electrolytes:

Li+(aq) + OH(aq) + H+(aq) + C2H3O2(aq)Li+(aq) + C2H3O2(aq) + H2O(l)

Net ionic – omit spectator ions:

H+(aq) + OH(aq) → H2O(l)

i)

Ionic – dissociate the electrolytes:

H+(aq) + Cl(aq) + NH3(aq) → NH4+(aq) + Cl(aq)

Net ionic – omit spectator ions:

H+(aq) + NH3(aq) → NH4+(aq)

j)

Ionic – dissociate the electrolytes:

6K+(aq) + 2PO43-(aq) + 3Ba2+(aq) + 6Cl(aq) → Ba3(PO4)2(s) + 6K+(aq) + 6Cl(aq)

Net ionic – omit spectator ions:

3Ba2+(aq) + 2PO43-(aq) → Ba3(PO4)2(s)

3.

Complete and balance each of the following molecular equations in aqueous solution:

a) AgNO3(aq) + CaCl2(aq) →

b) HNO3(aq) + Na2CO3(aq) →

c) K2CO3(s) + HClO3(aq) →

d) Al(OH)3(s) + HBr(aq) →

e) H3PO4(aq) + Fe(OH)3(s) →

f) BaCl2(aq) + K2SO4(aq) →

a)

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

b)

b) 2HNO3(aq) + Na2CO3(aq) → 2NaNO3(aq) + CO2(g)+ H2O(l)

c)

c) K2CO3(s) + HClO3(aq) → 2KClO3(aq) + CO2(g)+ H2O(l)

d)

d) Al(OH)3(s) + 3HBr(aq) → AlBr3(aq) + 3H2O(l)

e)

e) H3PO4(aq) + Fe(OH)3(s) → FePO4(s) + 3H2O(l)

f)

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

Solution

We predict the products by combining the cation from each reactant with the anion from the other. To obtain correct formulas, make sure you know the charge of the cation and the formula together with the charge of the anion. Adjust the subscript based on the charges and remember, the overall charge of the compound should balance to zero.

So, let’s make a list of steps for predicting the products of ionic reactions.

1) Identify the ions and write their charge on top

1a) Combine the cations and anions

2) Exchange the charges as subscripts

2a) Balance the equation and omit the charges

_____________________________________________________________________________

a) AgNO3(aq) + PbCl2(aq) →

1) Combine the cations and anions:

2) Balance the equation and omit the charges:

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

b) HNO3(aq) + Na2CO3(aq) →

1) Combine the cations and anions:

2) Balance the equation and omit the charges:

Remember that carbonic acid is unstable and decomposes immediately to CO2(g)+ H2O(l):

H2CO3(aq) → CO2(g)+ H2O(l)

So, in the final/corrected version of the equation, we write CO2 and H2O and not H2COwhich we use here to demonstrate how the combination of ions occurs.

2HNO3(aq) + Na2CO3(aq) → 2NaNO3(aq) + CO2(g)+ H2O(l)

c) K2CO3(s) + HClO3(aq) →

1) Combine the cations and anions:

2) Balance the equation and omit the charges:

Remember that carbonic acid is unstable and decomposes immediately to CO2(g)+ H2O(l):

H2CO3(aq) → CO2(g)+ H2O(l)

So, in the final/corrected version of the equation, we write CO2 and H2O and not H2COwhich we use here to demonstrate how the combination of ions occurs.

K2CO3(s) + HClO3(aq) → 2KClO3(aq) + CO2(g)+ H2O(l)

d) Al(OH)3(s) + HBr(aq) →

1) Combine the cations and anions:

2) Balance the equation and omit the charges:

Al(OH)3(s) + 3HBr(aq) → AlBr3(aq) + 3H2O(l)

Notice that HOH is a result of combining H+ and OH, and it is the same as H2O.

e) H3PO4(aq) + Fe(OH)3(s) →

1) Combine the cations and anions:

2) Balance the equation and omit the charges:

H3PO4(aq) + Fe(OH)3(s) → FePO4(s) + 3H2O(l)

f) BaCl2(aq) + K2SO4(aq) →

1) Combine the cations and anions:

2) Balance the equation and omit the charges:

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

4.

Calculate the molarity of each ion in the following solutions:

a) 0.150 M NaNO3
b) 0.250 M K2SO4
c) 0.280 M Ca(NO3)2
d) 0.350 M AlCl3

a)

0.150 M Naand 0.150 M NO3

b)

0.500 M K+ 0.250 M SO42-

c)

0.280 M Ca2+ and 0.560 M NO3

d)

0.350 M Al3+ 1.05 M  Cl

Solution

a) 0.150 M NaNO3

Each molecule/mole of NaNOcontains one/one mole of Naand one/one mole NO3 ion. Therefore, 0.150 M NaNOwill produce the same concentration of each ion.

You can also write the dissociation equation to see the mol/molar ratio:

NaNO3(aq) → Na+(aq) + NO3(aq)

Therefore, the concentration will be 0.150 M Naand 0.150 M NO3.

b) 0.250 M K2SO4

Each molecule/mole of K2SO4 contains two/two moles of Kand one/one mole SO42- ion. Therefore, 0.250 M K2SO4 will produce two times more K+ ions and the same concentration of SO42- ions.

You can also write the dissociation equation to see the mol/molar ratio:

K2SO4(aq) → 2K+(aq) + SO42-(aq)

Therefore, the concentration will be 0.500 M K+ 0.250 M SO42-.

c) 0.280 M Ca(NO3)2

Each molecule/mole of Ca(NO3)2 contains two/two moles of NO3 and one/one mole Ca2+ ion. Therefore, 0.280 M Ca(NO3)2 will produce two times more NO3 ions and the same concentration of Ca2+ ions.

You can also write the dissociation equation to see the mol/molar ratio:

Ca(NO3)2(aq) → Ca2+(aq) + 2NO3(aq)

Therefore, the concentration will be 0.280 M Ca2+ and 0.560 M NO3.

d) 0.350 M AlCl3

Each molecule/mole of AlCl3 contains three/three moles of Cl and one/one mole Al3+ ion. Therefore, 0.350 M AlCl3 will produce three times more Cl ions and the same concentration of Al3+ ions.

You can also write the dissociation equation to see the mol/molar ratio:

AlCl3(aq) → Al3+(aq) + 3Cl(aq)

Therefore, the concentration will be 0.350 M Al3+ 1.05 M  Cl.

5.

Which solution contains more chloride ions assuming complete solubility of salts?

a) 0.140 M NaCl or 0.060 M AlCl3
b) 0.250 M BaCl2 or 0.250 M MgCl2
c) 0.250 M MgCl2 or 0.450 M KCl

a)

0.060 M AlCl3

b)

The same concentration of Cl

c)

250 M MgCl2

Solution

a) 140 M NaCl and 0.060 M AlCl3

Each molecule/mole of NaCl contains one/one mole of Cl ion. Therefore, 0.140 M NaCl will produce the same concentration of Cl ion.

M(Cl) = 1 x 0.140 M = 0.140 M

On the other hand, 0.060 M AlCl3 will produce a three times higher concentration of Cl ion because there are three Cl ions in each molecule of AlCl3.

M(Cl) = 3 x 0.060 M = 0.180 M

b) 250 M BaCl2 or 0.250 M MgCl2

Each salt contains 2 ions of Cl, and therefore, the concentration of Clwill be twice that of each salt. Because the concentration of the salts are equal, they will produce the same amount of Clions.

M(Cl) in BaCl2 solution = 2 x 0.250 M = 0.500 M

M(Cl) in MgCl2 solution = 2 x 0.250 M = 0.500 M

c) 250 M MgCl2 or 0.450 M KCl

MgCl2 contains 2 ions of Cl, and therefore, the concentration of Clwill be twice that of each salt.

M(Cl) = 2 x 0.250 M = 0.500 M

Each molecule/mole of KCl contains one/one mole of Cl ion. Therefore, 0.450 M KCl will produce the same concentration of Cl ion.

M(Cl) = 1 x 0.450 M = 0.450 M

So, a solution with 0.250 M MgCl2 contains a higher concentration of Cl ions than 0.450 M KCl.

6.

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.

7.

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.

8.

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:

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

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

9.

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:

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

10.

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:

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

11.

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:

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

12.

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.

13.

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:

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

14.

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:

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

15.

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

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