Name _____________________________________
Solution Stoichiometry Worksheet
Solve the following solutions Stoichiometry problems:
1. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added
to 100. mL of 0.400 M potassium chromate?
2 AgNO
3(aq)
+ K
2
CrO
4(aq)
Ag
2
CrO
4(s)
+ 2 KNO
3(aq)
0.150 L AgNO
3
0.500 moles AgNO
3
1 moles Ag
2
CrO
4
331.74 g Ag
2
CrO
4
= 12.4 g Ag
2
CrO
4
1 L
2 moles AgNO
3
1 moles Ag
2
CrO
4
0.100 L K
2
CrO
4
0.400 moles K
2
CrO
4
1 moles Ag
2
CrO
4
331.74 g Ag
2
CrO
4
= 13.3 g Ag
2
CrO
4
1 L
1 moles K
2
CrO
4
1 moles Ag
2
CrO
4
2. How many mL of 0.280 M barium nitrate are required to precipitate as barium sulfate all the sulfate
ions from 25.0 mL of 0.350 M aluminum sulfate? (93.8 mL barium nitrate)
3 Ba(NO
3
)
2(aq)
+ Al
2
(SO
4
)
3(aq)
3 BaSO
4(s)
+ 2 Al(NO
3
)
3(aq)
0.350 moles Al
2
(SO
4
)
3
3 moles Ba(NO
3
)
2
1 L
= 0.0938 L Ba(NO
3
)
2
1 L
1 moles Al
2
(SO
4
)
3
0.280 moles Ba(NO
3
)
2
3. 25.0 mL of 0.350 M NaOH are added to 45.0 mL of 0.125 M copper (II) sulfate. How many grams of
copper (II) hydroxide will precipitate?
2 NaOH
(aq)
+ CuSO
4(aq)
Cu(OH)
2(s)
+ Na
2
SO
4(aq)
0.0250 L NaOH
0.350 moles NaOH
1 moles Cu(OH)
2
97.57 g Cu(OH)
2
= 0.427 g Cu(OH)
2
1 L NaOH
2 moles NaOH
1 mole Cu(OH)
2
0.0450 L CuSO
4
0.125 moles CuSO
4
1 moles Cu(OH)
2
97.57 g Cu(OH)
2
= 0.549 g Cu(OH)
2
1 L NaOH
1 moles CuSO
4
1 mole Cu(OH)
2
4. What volume of 0.415 M silver nitrate will be required to precipitate as silver bromide all the bromide
ion in 35.0 mL of 0.128 M calcium bromide?
2 AgNO
3(aq)
+ CaBr
2(aq)
Ca(NO
3
)
2(aq)
+ 2 AgBr
(s)
0.0350 L CaBr
2
0.128 moles CaBr
2
2 moles AgNO
3
1 L AgNO
3
= 0.0216 L AgNO
3
1 L CaBr
2
1 moles CaBr
2
0.415 mole AgNO
3
5. What volume of 0.496 M HCl is required to neutralize 20.0 mL of 0.809 M sodium hydroxide?
HCl(aq) + NaOH(aq) NaCl(aq) + H(OH)(l)
0.0200 L NaOH
0.809 mole NaOH
1 mole HCl
1 L HCl
= 0.0326 L HCl
1 L NaOH
1 mole NaOH
0.496 mole HCl
6. How many mL of 0.715 M HCl is required to neutralize 1.25 grams of sodium carbonate? (producing
carbonic acid)
2 HCl
(aq)
+ Na
2
CO
3(s)
2 NaCl
(aq)
+ H
2
CO
3(aq)
1.25 g Na
2
CO
3
1 mole Na
2
CO
3
2 mole HCl
1 L HCl
= 0.0330 L HCl
105.99 g Na
2
CO
3
1 mole Na
2
CO
3
0.715 mole HCl
7. What minimum number of grams of oxalic acid monohydrate, H
2
C
2
O
4
• H
2
O, would you specify for a
titration of no fewer than 15.0 mL of 0.100 M NaOH? Both of the hydrogen’s from oxalic acid are
replaceable in this reaction.
H
2
C
2
O
4
• H
2
O(aq) + 2 NaOH(aq) Na
2
C
2
O
4
• H
2
O(aq) + 2 H(OH)(l)
0.0150 L NaOH
0.100 mole NaOH
1 mole H
2
C
2
O
4
• H
2
O
108.06 g H
2
C
2
O
4
• H
2
O
= 0.0810 g H
2
C
2
O
4
• H
2
O
1 L NaOH
2 mole NaOH
1 mole H
2
C
2
O
4
• H
2
O
8. How many grams of magnesium hydroxide will precipitate if 25.0 mL of 0.235 M magnesium nitrate
are combined with 30.0 mL of 0.260 M potassium hydroxide?
Mg(NO
3
)
2(aq)
+ 2 KOH 2 KNO
3(aq)
+ Mg(OH)
2(s)
0.0250 L Mg(NO
3
)
2
0.235 mole Mg(NO
3
)
2
1 mole Mg(OH)
2
58.33 g Mg(OH)
2
= 0.343 Mg(OH)
2
1 L Mg(NO
3
)
2
1 mole Mg(NO
3
)
2
1 mole Mg(OH)
2
0.0300 L KOH
0.260 mole KOH
1 mole Mg(OH)
2
58.33 g Mg(OH)
2
= 0.227 g Mg(OH)
2
1 L KOH
2 mole KOH
1 mole Mg(OH)
2
9. 60.0 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 M lead (II) nitrate.
How many grams of lead (II) iodide will precipitate?
2 KI
(aq)
+ Pb(NO
3
)
2(aq)
2 KNO
3(aq)
+ PbI
2(s)
0.0600 L KI
0.322 mole KI
1 mole PbI
2
461.00 g PbI
2
= 4.45 g PbI
2
1 L KI
2 mole KI
1 mole PbI
2
0.0200 L Pb(NO
3
)
2
0.530 mole Pb(NO
3
)
2
1 mole PbI
2
461.00 g PbI
2
= 4.89 g PbI
2
1 L Pb(NO
3
)2
1 mole Pb(NO
3
)2
1 mole PbI
2