Name
Period
Date
Pre-AP Chemistry
Homework: Calculating with Chemical Equations
[A] Stoichiometry Problems
1. ___________ Calculate the mass of magnesium oxide formed when 0.52 g of magnesium is burned according
to the following equation:
2 Mg(s) + O2(g)  2 MgO(s)
2. ___________ Calculate the mass of hydrogen gas that will be released when 3.0 g of sodium reacts with water.
2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)
3. ___________ What mass of silver will be precipitated when 7.32g of zinc reacts completely with silver nitrate?
Zn(s) + 2 AgNO3(aq)  Zn(NO3)2(aq) + 2 Ag(s)
4. ___________ Potassium metal reacts with hydrochloric acid to produce aqueous potassium chloride and
hydrogen gas. How many grams of potassium are required to produce 5.00 g of hydrogen gas?
5. ___________ Ethanol burns according to the following equation. If 655 g of water is produced, what mass of
ethanol is burned?
C2H5OH(aq) + 3O2(g)  2CO2(g) + 3H2O(g)
6. ___________ How many grams of potassium chlorate, KClO3, must be decomposed to produce oxygen and
71.8 g of potassium chloride?
[B] Limiting Reactant Problems
1. Calcium hydroxide, used to neutralize acid spills, reacts with hydrochloric acid according to the following
equation:
Ca(OH)2 + 2HCl CaCl2 + 2H2O
a. ___________ If you have spilled 6.3 mol of HCl and put 2.8 mol of Ca(OH) 2 on it, which substance is the
limiting reactant?
b. ___________ How many moles of the excess reactant remain?
2. Aluminum oxidizes according to the following equation:
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4Al + 3O2  2Al2O3
a. ___________ Powdered Al (0.048 mol) is placed into a container containing 0.030 mol O 2. What is limiting
reactant?
b. ___________ How many moles of the excess reactant remain?
[C] Percent Yield Problems
1. ___________ Dichlorine monoxide, Cl2O is sometimes used as a powerful chlorinating agent in research. It
can be produced by passing chlorine gas over heated mercury (II) oxide according to the following equation. What
is the percent yield, if the quantity of the reactants is sufficient to produce 0.86g of Cl 2O but only 0.71 g is
obtained?
HgO + Cl2 HgCl2 + Cl2O
2. In the commercial production of the element arsenic, arsenic(III) oxide is heated with carbon, which reduces the
oxide to the metal according to the following equation:
2As2O3 + 3C 3CO2 + 4As
a. ___________ If 8.87g of As2O3 is used in the reaction and 5.33 g of As is produced, what is the percent yield?
b. ___________ If 67 g of carbon is used up in a different reaction and 425g of As is produced, calculate the
percent yield of this reaction.
3. ___________ Huge quantities of sulfur dioxide are produced from zinc sulfide by means of the following
reaction. If the typical yield is 86.78%, how much SO2 should be expected if 4897g of ZnS are used?
2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g)
4. ___________ A process by which zirconium metal can be produced from the mineral zirconium (IV)
orthosilicate, ZrSiO4, starts by reacting it with chlorine gas to form zirconium (IV) chloride. What mass of ZrCl4
can be produced if 862g of ZrSiO4 and 950.g of Cl2 are available? (You must first determine limiting reactant).
ZrSiO4 + 2Cl2  ZrCl4 +SiO2 + O2
Homework: Calculating with Chemical Equations
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2/8/16
Name
Period
Date
Pre-AP Chemistry
Homework: Calculating with Chemical Equations
[A] Stoichiometry Problems
1. ___________ Calculate the mass of magnesium oxide formed when 0.52 g of magnesium is burned according
to the following equation:
2 Mg(s) + O2(g)  2 MgO(s)
0.52g Mg ×
= 0.0214 mole Mg
×
=0.0214 mole MgO
×
= 0.8626 = 0.86 grams of MgO
2. ___________ Calculate the mass of hydrogen gas that will be released when 3.0 g of sodium reacts with water.
2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)
3g Na ×
= 0.1305 mole Na×
=0.0652 mole H2
×
= 0.1315 grams H2 = 0.13 grams of H2
3. ___________ What mass of silver will be precipitated when 7.32g of zinc reacts completely with silver nitrate?
Zn(s) + 2 AgNO3(aq)  Zn(NO3)2(aq) + 2 Ag(s)
7.32 g Zn ×
= 0.1119 mole Zn
×
=0.2238 mole Ag
×
=24.1501 = 24.2 grams of Ag
4. ___________ Potassium metal reacts with hydrochloric acid to produce aqueous potassium chloride and
hydrogen gas. How many grams of potassium are required to produce 5.00 g of hydrogen gas?
2K + 2HCl  2KCl + H2
5.00g H2 ×
=2.4802 mole H2
×
=4.9603 mole K
×
= 193.4524 = 193 grams K
5. ___________ Ethanol burns according to the following equation. If 655 g of water is produced, what mass of
ethanol is burned?
C2H5OH(aq) + 3O2(g)  2CO2(g) + 3H2O(g)
655 g H2O ×
= 36.3566 mole H2O ×
=12.1189 mole C2H5OH ×
= 558.2915 = 558 g of
ethanol
6. ___________ How many grams of potassium chlorate, KClO3, must be decomposed to produce oxygen and
71.8 g of potassium chloride?
2KClO3  2KCl + 3O2
71.8g KCl ×
= 0.9631 mole KCl
×
=0.9631 mole KClO3
×
= 118.0294 = 118 grams KClO3
[B] Limiting Reactant Problems
1. Calcium hydroxide, used to neutralize acid spills, reacts with hydrochloric acid according to the following
equation:
Ca(OH)2 + 2HCl CaCl2 + 2H2O
a. ___________ If you have spilled 6.3 mol of HCl and put 2.8 mol of Ca(OH)2 on it, which substance is the
limiting reactant?
6.3 mole HCl ×
= 3.15 mole of Ca(OH)2 needed. You have only 2.8 mole which isn’t enough. Ca(OH)2 is the limiting
reactant.
b. ___________ How many moles of the excess reactant remain?
Use the limiting reactant to do the calculations
Homework: Calculating with Chemical Equations
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2.8 mole Ca(OH)2 ×
=5.6 mole of HCl needed. You have 6.3 (this also confirms that the HCl is in excess)…6.3 – 5.8
actually used = 0.7 mole of HCl remaining
2. Aluminum oxidizes according to the following equation:
4Al + 3O2  2Al2O3
a. ___________ Powdered Al (0.048 mol) is placed into a container containing 0.030 mol O 2. What is limiting
reactant?
0.048 mole Al ×
= 0.036 mole of O2 needed. You only have 0.030, so the O2 is the limiting reactant (there isn’t enough
for this much Al)
b. ___________ How many moles of the excess reactant remain?
Use limiting reactant to get the amounts actually used
0.030 O2 ×
=0.04 mole of Al actually needed. 0.048 available – 0.040 needed = 0.008 mole of Al remaining.
[C] Percent Yield Problems
1. ___________ Dichlorine monoxide, Cl2O is sometimes used as a powerful chlorinating agent in research. It
can be produced by passing chlorine gas over heated mercury (II) oxide according to the following equation. What
is the percent yield, if the quantity of the reactants is sufficient to produce 0.86g of Cl 2O but only 0.71 g is
obtained?
HgO + Cl2 HgCl2 + Cl2O
This is the simplest form of a percent yield problem because it already tells you much was actually produced and how much could
have been produced. It’s basically a comparison of these two values.

=82.5581=83%. This means that they obtained 83% of the ideal or expected mass.
2. In the commercial production of the element arsenic, arsenic(III) oxide is heated with carbon, which reduces the
oxide to the metal according to the following equation:
2As2O3 + 3C 3CO2 + 4As
a. ___________ If 8.87g of As2O3 is used in the reaction and 5.33 g of As is produced, what is the percent yield?
This is a harder version because you have to first calculate how much should have been produced. In the previous problem, they
just told you that 0.86 was expected. Here you have to do a whole stoichiometry problem to get to the expected value.
8.87 As2O3 ×
=0.0448 mole reacted ×
= 0.0897 mole of As expected ×
=6.718 = 6.72 grams, ideally
Now it’s like the first problem:
=79.3396 = 79% yield
b. ___________ If 67 g of carbon is used up in a different reaction and 425g of As is produced, calculate the
percent yield of this reaction.
67g C ×
= 5.5787 mole C
×
=7.4382 mole As
×
=557.2734 grams of As, theoretically
=76.2642 = 76.3%
3. ___________ Huge quantities of sulfur dioxide are produced from zinc sulfide by means of the following
reaction. If the typical yield is 86.78%, how much SO2 should be expected if 4897g of ZnS are used?
2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g)
In this problem they tell you what the percent yield is and how much you started with. You have to figure out, using this starting
gram amount, how much could be made (as if it were 100% yield) and then take 86.78% of this ideal amount.
4897 g ZnS ×
=50.2359 moles ZnS ×
= 50.2359 moles SO2 expected ×
= 3218.617 grams of SO2
expected
=86.78%  x=
=2793.1159=2793 grams if SO2, if the yield is only 86.78%
4. ___________ A process by which zirconium metal can be produced from the mineral zirconium (IV)
orthosilicate, ZrSiO4, starts by reacting it with chlorine gas to form zirconium (IV) chloride. What mass of ZrCl4
can be produced if 862g of ZrSiO4 and 950.g of Cl2 are available? (You must first determine limiting reactant).
ZrSiO4 + 2Cl2  ZrCl4 +SiO2 + O2
Homework: Calculating with Chemical Equations
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862 g ZrSiO4
=4.7024 mole of ZrSiO4 present ×
=9.4048 mole Cl2 needed ×
=666.8027 g Cl2
needed. We have way too much (actually have 950; so Cl2 is in excess); You can use the ZrSiO4 value to predict how much ZrCl4
will be produced.
4.7024 mole of ZrSiO4 present ×
=4.7024 mole of ZrCl4 will be produced, ideally ×
=1095.7532=1010. g or
1.10x103 g ZrCl4
Homework: Calculating with Chemical Equations
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Answers: (1) 0.86 g (2) 1.98 g (3) 0.13 g (4) 13.4 g (5) 24.2 g (6) 194 g (7) 558 g (8) 118 g
1a) Ca(OH)2 is the limiting reactant; 1b)0.7 mol HCl will be left over; 2a)O2 is the limiting reactant; 2b)0.008 mol Al will be left
over.
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