Chapter 11
Name ___________________________
Period _____
Study Packet: Quantifying Chemical Compounds
Answers Only
Set I. Review and Molar Mass
1. Determine the moles of aluminum in 23 g of Al.
G: 23 g Al
W: moles Al
R: 1 mol al = 26.982 g Al
23 g Al
1 mole Al
x
= 0.852 mol Al
1
26.982 g Al
0.852 mol Al
2. Determine the number of atoms of helium in 23 moles of helium.
G: 23 mol He
W: atoms of He
R: 1 mole = 6.02 x 1023atoms
23 mol He x 6.02 x 1023atoms = 1.38 x 1025 atoms He
1
1 mol
1.38 x 1025 mc He
25
1.38 x 10 atoms He
1.38 x 1025 mc He
mol Al
3. Complete the table below
Particle Type
Name
Formula
(mc, f.un, or atoms)
Molar Mass (g/mol)
potassium nitrate
f.un
KNO3
calcium chloride
f.un
CaCl2
110.986
Magnesium phosphate
f.un
Mg3(PO4)2
262.855
sodium nitride
f.un
Na3N
82.977
Magnesium carbonate
f.un
MgCO3
84.313
Silicon dioxide
Mc
SiO2
60.084
carbon tetrachloride
Mc
CCl4
153.823
oxygen
mc
O2
31.998
potassium chlorate
f.un
KClO3
122.548
iron (II) sulfide
f.un
FeS
87.913
Adv Chem
101.102
1
Set II. Usin’ Mol’r Mass
Work the following problems on separate paper. Use the molar masses from Set I as a quick reference.
1. Determine the mass of 3.56 x 1024 formula units of sodium nitride.
G: 3.56 x 1024 f.un Na3N
W: mass Na3N
R: 1 mol = 6.02 x 1023 f.un
1 mol Na3N = 82.977 g Na3N
3.56 x 1024 f.un Na3N
1
x
1 mol
x 82.977 g Na3N
23
6.02 x 10 f.un
1 mol Na3N
=
491 g Na3N
2. Determine the mass of 126 moles of carbon tetrachloride.
G: 126 mol CCl4
W: mass CCl4
R: 1 mol CCl4 = 153.823 g
4
126 mol CCl4 x 153.823 g CCl4 = 1.94 x 10 g CCl4
1
1 mol CCl4
3. Determine the number of moles in 5.6 mg of silicon dioxide.
G: 5.6 mg SiO2
W: moles SiO2
R: 1 mole SiO2 = 60.084 g SiO2
1 g = 1000 mg
5.6 mg SiO2
1
x
1g
x
1000 mg
9.32 x 10-5 mol SiO2
1 mole SiO2 =
60.084 g SiO2
4. Determine the number of molecules in 5 lbs of sand (silicon dioxide).
G: 5 lb SiO2
W: mc SiO2
R: 1 lb = 454 g
1 mol SiO2 = 60.084 g SiO2
1 mol = 6.02 x 1023 mc
5 lb SiO2 x 454 g x 1 mol SiO2 x 6.02 x 1023 mc = 2.27 x
1
1 lb
60.084 g SiO2
1 mol
5. Determine the mass (in kg) of 22.3 moles of magnesium phosphate.
G: 22.3 mol Mg3(PO4)2
W: mass(kg) Mg3(PO4)2
R: 1 kg = 1000 g
1 mol Mg3(PO4)2 = 262.855 g Mg3(PO4)2
22.3 mol Mg3(PO4)2
1
Adv Chem
x
262.855 g Mg3(PO4)2
1 mol Mg3(PO4)2
x
1 kg = 5.86
1000 g
1025 mc SiO2
kg Mg3(PO4)2
2
6. Determine the number of formula units in 122 g of iron(III) oxide
G: 122 g Fe2O3
W: f.un Fe2O3
R: 1 mol = 6.02 x 1023 f.un
1 mol Fe2O3 =159.691 g Fe2O3
122 g Fe2O3
1
x
1 mol Fe2O3
159.691 g Fe2O3
x
6.02 x 1023 f.un
1 mol
=
4.60 x 1023 f.un Fe2O3
7. Determine the number of moles in 8.6 x1023 formula units of calcium chloride.
G: 8.62 x 1023 f.un CaCl2
W: mol CaCl2
R: 1 mol = 6.02 x 1023 f.un
8.62 x 1023 f.un CaCl2
1
x
1 mol
6.02 x 1023 f.un
=
1.43 mol CaCl2
Set III. Per Comp
Work the following problems on separate paper. Use the molar masses from Set I as a quick reference.
1. Determine the % composition of each element in…
a)
potassium nitrate
K: 39.098 g/mol
101.102 g/mol
38.7 %
N:
13.9 %
14.007 g/mol
101.102 g/mol
O:
3 x (15.999g/mol) 47.5 %
101.102 g/mol
b)
carbon tetrachloride
C: 12.011 g/mol
153.823 g/mol
7.81 %
Cl: 4 x (35.453 g/mol) 92.2 %
153.834 g/mol
c)
KMnO4
K:
Mn:
O:
d)
39.098 g/mol
158.032 g/mol
54.938 g/mol
158.032 g/mol
24.6 %
34.8 %
4 X (15.999g/mol) 40.5 %
158.032 g/mol
Na2CO3
Na: 2 x (22.990g/mol) 43.4%
105.988 g/mol
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3
C:
12.011 g/mol
105.988 g/mol
11.3%
O:
3 x (15.999g/mol)
45.2 %
105.988 g/mol
2. Determine the mass of potassium in a 30.0 g of potassium nitrate.
XgK
answer: 11.6 g K
= 39.098 g/mol K
30 g KNO3
101.102 g/mol KNO3
3. Determine the mass of bromine in 50.0 g of potassium bromide.
X g Br =
79.904 g/mol Br
answer: 33.6 g Br
50 g KBr
119.002 g/mol KBr
QUIZ TOMORROW- STUDY!!!
Stoichiometry
Set IV. Conversion Factors
Consider the following reaction:
1. 3 Hg(OH)2 + 2 H3PO4 ---> Hg3(PO4)2 + 6 H2O
a) How many moles of reactants are there? 5 moles of reactant
b) How many moles of products are there? 7 moles of product
c) Identify all of the molecular compounds. H2O
d) Identify all of the ionic compounds. Hg3(PO4)2, H3PO4, Hg(OH)2
e) If 10 formula units of Hg3(PO4)2 are produced in a reaction, how many molecules of H2O are
produced? 60 molecules of water
Balance the following equations
Write out all of the possible mole ratios
2. _2_SO2 + __O2  _2_SO3
2. 2 mol SO2 = 1 mole O2, 1mole SO2 = 1
mole SO3, 1 mole O2 = 2 mole SO3
3. __PCl3 + __Cl2  __PCl5
3. Everything is 1:1
4. 4 mol NH3 = 3 mol O2, 2 mol NH3 = 1 mol
4. _4_NH3 + _3_O2  _2_N2 + _6_H2O
5. __Fe2O3 + _3_CO  _2_ Fe + _3_CO2
Adv Chem
N2, 2 mol NH3 = 3 mol H20, 3 mol O2 = 2 mol
N2, 1 mol O2 = 2 mol H2O 1 mol N2 = 3 mol
H2O
5. 1 mol Fe2O3= 3 mol CO, 1 mol Fe2O3= 2
mol Fe, 1 mol Fe2O3=3 mol CO2, 3 mol CO = 2
mol Fe, 1 mol CO = 1 mol CO2, 2 mol Fe = 3
mol CO2
4
Set V. Mole to mole and mole to mass
On a separate sheet of paper balance the equations for each problem and solve the questions. Don’t
forget to make sure the reaction is balanced!
1. “sulfur burns in oxygen gas to form sulfur dioxide”
S + O2 SO2
a. How many moles of sulfur must be burned to give 100.0 g of SO2 ?
G: 100.0 g SO2
W: moles S
R: 1 mol S = 1 mol SO2
1 mol SO2 = 64.064 g SO2
100 g SO2 x 1 mol SO2
1
64.064 g SO2
x
1 mol S
1 mol SO2
=
1.56 mol S
b. How many moles of sulfur must be burned to completely react with 10 moles of oxygen?
G: 10 mol O2
W: mol S
R: 1 mol S = 1 mol O2
10 mol O2 x 1 mol S = 10
1
1 mol O2
mol S
c. If there are 32 moles of oxygen, how many molecules of oxygen are there?
G: 32 mol O2
W: mc O2
R: 1 mol = 6.02 x 1023 mc
32 mol O2
1
x
6.02 x 1023 mc
1 mol
=
1.93 x 1025 mc O2
2. BaO + 2 HNO3  Ba(NO3)2 + H2O
a. How many mg of Ba(NO3)2 can be formed from 196.0 moles of HNO3?
G: 196.0 moles HNO3
W: mg of Ba(NO3)2
R: 1g = 1000 mg
1 mole Ba(NO3)2 = 261.338 g Ba(NO3)2
1 mole Ba(NO3)2 = 2 moles HNO3
196.0 moles HNO3 x 1 mole Ba(NO3)2 x 261.338 g Ba(NO3)2 x 1000 mg = 2.56 x 107 mg Ba(NO3)2
1
2 moles HNO3
1 mole Ba(NO3)2
1g
b. If 7.45 moles of water are formed during this reaction, how many moles of HNO3 were used?
G: 7.45 mol water
W: moles HNO3
R: 1 mole H2O = 2 HNO3
7.45 mol H2O x 2 mol HNO3
1
1 mole H2O
Adv Chem
14.9 mol HNO3
=
5
3. Cu + 3AgNO3  Cu(NO3)3 + 3Ag
a. How many moles of AgNO3 are needed to produce 9.85 mol of silver.
G: 9.85 mol of silver
W: moles of AgNO3
R: 3 mol Ag = 3 mol AgNO3
9.85 mol Ag x 3 mol AgNO3 = 9.85 mol AgNO3
1
3 mol Ag
b. How many moles of copper produce 5 kg of silver?
G: 5 kg of silver
W: moles of copper
R: 1 kg = 1000 g
1 mole Cu = 3 mole Ag
1 mole Ag = 107.868 g Ag
5 kg Ag x 1000 g
1
x
1 kg
1 mole Ag
107.868 g Ag
x
1 mole Cu = 15.5
3 mole Ag
mole Cu
5. 2 NaCl + Zn(NO3)2  ZnCl2 + 2 NaNO3
a. 6 moles NaCl should produce how many moles of ZnCl2?
G: 6 moles NaCl
W: moles of ZnCl2
R: 2 moles NaCl = 1 mole ZnCl2
6 moles NaCl x 1 mole ZnCl2 = 3 mol ZnCl2
1
2 moles NaCl
b. How many moles of ZnCl2 can be produced from 107.0 g of Zn(NO3)2?
G: 107.0 g of Zn(NO3)2
W: moles of ZnCl2
R: 1 mole Zn(NO3)2 = 1 mole ZnCl2
1 mole Zn(NO3)2 = 189.398 g Zn(NO3)2
107.0 g Zn(NO3)2
1
x
1 mole Zn(NO3)2
189.398 g Zn(NO3)2
x
1 mole ZnCl2 = 0.565
1 mole Zn(NO3)2
mol ZnCl2
Extra Practice
NaOH + Al  Na3AlO3 + H2
a. How many moles of hydrogen will be produced if 2 moles of NaOH react will excess Al?
b. How many moles of hydrogen can be prepared from 1 gram of aluminum?
c. If 3 moles of Na3AlO3 are produced, how many moles of Al were present before the reaction if the
aluminum all reacted?
d. How many kg of Na3AlO3 can be formed from 165.0 moles of sodium hydroxide?
e. How many moles of NaOH are required to produce 3 g of hydrogen?
Set VI. Mass to mass and more
On a separate sheet of paper balance the equations for each problem and solve the questions.
1. NaCl + AgNO3  AgCl + NaNO3
a. 78.00 g of NaCl should produce how many grams of AgCl?
G: 78.00 g NaCl
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W: g AgCl
R: 1 mol NaCl = 1 mol AgCl
1 mol NaCl = 58.443g NaCl
1 mol AgCl = 143.321 g AgCl
78.00g NaCl x 1 mol NaCl x 1 mol AgCl x 143.321 g AgCl = 191.28g AgCl
1
58.443 g NaCl 1 mol NaCl
1 mol AgCl
b. 78.00 g of NaCl are required to react completely with how many grams of silver nitrate?
G: 78.00 g NaCl
W: g AgNO3
R: 1 mol NaCl = 1 mol AgNO3
1 mol NaCl = 58.442 g NaCl
1 mol AgNO3 = 169.872 g AgNO3
78.00g NaCl x 1 mol NaCl
x 1 mol AgNO3 x 169.872g AgNO3 = 227g AgNO3
1
58.443 g NaCl
1 mol NaCl
1 mol AgNO3
c. If 50.0 grams of silver nitrate are consumed in a reaction, how many grams of sodium nitrate
are produced with it?
G: 50.0 g AgNO3
W: g NaNO3
R: 1 mol AgNO3 = 1 mol NaNO3
1 mol AgNO3 = 169.872 g AgNO3
1 mol NaNO3 = 84.994 g AgNO3
50.0g AgNO3 x 1 mol AgNO3
x 1 mol NaNO3 x 84.994g NaNO3 = 25.01g NaNO3
1
169.872g AgNO3
1 mol AgNO3
1 mol NaNO3
d. How many moles of sodium nitrate will form from 4 moles of sodium chloride?
G: 4 mol NaCl
W: mol AgNO3
R: 1 mol NaCl = 1 mol AgNO3
4 mol NaCl x 1 mol AgNO3 = 4 mol AgNO3
1
1 mol NaCl
e. How many grams of silver chloride will form from 2 moles of sodium chloride?
G: 2 mol NaCl
W: g AgCl
R: 1 mol NaCl = 58.442 g NaCl
1 mol AgCl = 143.321 g AgCl
2 mol NaCl x 1 mol AgCl x 143.321g AgCl = 287g Ag
1
1 mol NaCl
1 mol AgCl
f. If there are 60 g of silver nitrate how many moles of silver nitrate are there?
G: 60g AgNO3
W: 1 mol AgNO3 = 169.872 g AgNO3
60.0g AgNO3 x
1 mol AgNO3 = 0.350 mol AgNO3
1
169.872 g AgNO3
Adv Chem
7
2. 3 NaOH + Al  Na3AlO3 + 3 H2
a. How many moles of aluminum are required to produce 17.5 g of hydrogen?
b. How many kg of NaOH are required to react with 6.5 grams of Al ?
c. If 42.3 g of Na3AlO3 are produced in the reaction, how many g. of NaOH and Al are used?
d. How many moles of sodium hydroxide are in 76 g of sodium hydroxide?
G: 76 g NaOH
W: mol NaOH
R: 1 mol NaOH = 39.997g
76 g NaOH x
1 mol NaOH = 1.90 mol NaOH
1
39.997g NaOH
e. How many moles of hydrogen are released if 7 moles of aluminum are consumed?
G: 7 mol Al
W: mol H2
R: 2 mol Al = 3 mol H2
7 mol Al x 3 mol H2 = 10.5 mol H2
1
2 mol Al
Set VII. Molecules and formula units to mass and more
On a separate sheet of paper balance the equations for each problem and solve the questions.
1. Na2O + H2O  2NaOH
a. How many grams of NaOH are produced from 2.01 x 1023 molecules of water?
2.01 x 1023 mc H2O x
1 mol H2O
x 2 mol NaOH x 39.997g NaOH = 26.7g NaOH
1
6.02 x 1023 mc H2O
1 mol H2O
1 mol NaOH
b. How many formula units of Na2O are required to produce 1.60 grams of NaOH?
1.60g NaOH x 1 mol NaOH x 1 mol Na2O x 6.02 x 1023 fun Na2O = 1.20 x 1022 fun Na2O
1
39.997 NaOH
2 mol NaOH
1 mol Na2O
c. How many molecules of water are needed to react completely with 4.35 moles of sodium oxide?
4.35 mol Na2O x 1 mol H2O x 6.02 x 1023 fun Na2O = 2.61 x 1024 mc H2O
1
1 mol Na2O
1 mol Na2O
2. CH4 + 2O2  CO2 + 2H2O (when heated)
a. If 30.0 g of methane are combusted completely, how many milligrams of water are produced?
30.0 g CH4 x
1 mol CH4 x 2 mol H2O x 18.015 g H2O x 1000 mg = 6.73 x 104 mg H2O
1
16.043 g CH4 1 mol CH4
1 mol H2O
1g
b. How many grams of O2 are needed to completely combust with 4 moles of methane?
4 mol CH4 x 2 mol O2 x 31.998 g O2 = 255 g O2
1
1 mol CH4
1 mol O2
d. How many molecules of Carbon dioxide are formed from 2.90 x 1022 molecules of methane?
2.90 x 1022 mc CH4 x
1 mol CH4
x 1 mol CO2 x 6.02 x 1023 mc CO2 = 2.90 x 1022 mc
CO2
1
6.02 x 1023 mc CH4 1 mol CH4
1 mole CO2
3. Cu + 3AgNO3  Cu(NO3)3 + 3Ag
a. How many atoms of Ag are produced when 2.44 moles of Cu are consumed?
2.44 mol Cu x 3 mol Ag x 6.02 x 1023 atoms Ag = 4.41 x 1024 atoms Ag
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8
1
1 mol Cu
1 mol Ag
b. How many atoms of copper produce 4.22 x 1022 formula units of silver?
4.22 x 1023 fun Ag x
1 mol Ag
x 1 mol Cu x 6.02 x 1023 atoms Cu = 1.41 x 1022
atoms Cu
1
6.02 x 1023 fun Ag
3 mol Ag
1 mol Cu
Extra Practice
BaO + H2SO4  BaSO4 + H2O
a. How many formula units of BaSO4 are present if there are 43 g of BaSO4?
43 g BaSO4 x
1 mol BaSO4
x 6.02 x 1023 fun BaSO4 = 1.11 x 1023 fun BaSO4
1
233.392 g BaSO4
1 mol BaSO4
b. How many moles of water are formed from 6.04 x 1020 molecules of sulfuric acid?
6.04 x 1020 mc H2SO4 x
1 mol H2SO4
x 1 mol H2O
= 1.00 x 10-3 mol H2O
23
1
6.02 x 10 mc H2SO4
1 mol H2SO4
c. How many formula units of BaSO4 can be produced from 3.01 x 1023 formula units of BaO?
3.01 x 1023 fun BaO x 1 mol BaO
x 1 mol BaSO4 x 6.02 x 1023 fun BaSO4 = 3.01 x 1023 fun
BaO
1
6.02 x 1023 fun BaO 1 mol BaO
1 mol BaSO4
d. How many formula units of barium oxide are used if 26 g of water are produced?
26g H2O x
1 mol H2O
x 1 mol BaO x 6.02 x 1023 fun BaO = 8.69 x 1023 fun BaO
1
18.015g H2O
1 mol H2O
1 mol BaO
e. If 4 moles of water are formed how many formula units of Barium oxide were consumed?
4 mol H2O x 1 mol BaO x 6.02 x 1023 fun BaO = 2.41 x 1024 fun BaO
1
1 mol H2O
1 mol BaO
Set VIII. Avogadro’s Law
On a separate sheet of paper balance the equations for each problem and solve the questions.
1. N2(g) + H2(g)  NH3(g)
a. How many mol of N2 gas are needed to react with 100.0 L of H2 gas to produce NH3 gas?
G: 100L H2
W: mol N2
R: 1 mol H2 = 22.4 L H2
100.0 L H2 x 1 mol H2 x 1 mol N2 = 1.49 mol N2
1
22.4L H2
3 mol H2
b. How many liters of ammonia are produced when 89.6 L of H2 are used?
G: 89.6 L H2
W: L NH3
R: 1 mol H2 = 22.4 L H2
1 mol NH3 = 22.4 L NH3
2 mol NH3 = 3 mol H2
89.6 L H2 x 1 mol H2 x 2 mol NH3 x 22.4 L NH3 = 59.7 L NH3
1
22.4L H2
3 mol H2
1 mol NH3
OR
Adv Chem
89.6 L H2 x 2 L NH3 = 59.7 L NH3
1
3 L H2
9
c. If there are 9.78 x 1023 mc of nitrogen gas, how many liters of N2 are there?
G: 9.78 x 1023 mc N2
W: L N2
R: 1 mol N2 = 6.02 x 1023 mc N2
1 mol N2 = 22.4 L N2
9.78 x 1023 mc N2 x
1 mol N2
x 22.4 L N2
1
6.02 x 1023 mc N2 1 mol N2
= 36.4 L N2
d. If 9 liters of H2 are used, how many liters of ammonia are produced?
G: 9L H2
W: L NH3
R: 1 mol H2 = 22.4 L H2
OR 3L H2 = 2 L NH3
1 mol NH3 = 22.4L NH3
3 mol H2 = 2 mol NH3
9L H2 x 1 mol H2 x 2 mol NH3 x 22.4L NH3 = 6L NH3
1
22.4 L H2 3 mol H2
1 mol NH3
OR
9L H2 x 2L NH3 = 6L NH3
1
3L H2
e. If 12 moles of NH3 are produced, how many molecules of nitrogen were used?
G: 12 mol NH3
W: mc N2
R: 2 mol NH3 = 1 mol N2
1 mol N2 = 6.02 x 1023 mc N2
12 mol NH3 x 1 mol N2 x 6.02 x 1023 mc N2 = 3.61 x 1024 mc N2
1
2 mol NH3
1 mol N2
f. If 7 liters of ammonia are produced, how many liters of H2 were consumed?
G: 7 L NH3
W: L H2
R: 2 mol NH3 = 3 mol H2
OR 2 L NH3 = 3 L H2
1 mol NH3 = 22.4 L NH3
1 mol H2 = 22.4L H2
7L NH3 x 1 mol NH3 x 3 mol H2 x 22.4L H2 = 10.5L H2
1
22.4 L NH3 2 mol NH3 1 mol H2
OR
7L NH3 x 3 L H2 = 10.5L H2
1
2 L NH3
g. If 13 liters of hydrogen were used, how many molecules of nitrogen reacted?
G: 13L H2
W: mc N2
R: 3 mol H2 = 1 mol N2
1 mol H2 = 22.4 L H2
1 mol N2 = 6.02 x 1023 mc N2
13 L H2 x 1 mol H2 x 1 mol N2 x 6.02 x 1023 mc N2 = 1.16 x 1023 mc N2
1
22.4L H2
3 mol H2
1 mol N2
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Set IX. Theoretical yield and percent yield.
On a separate sheet of paper balance the equations for each problem and solve the questions.
1. Calculate the percent yield based on the provided data:
a. theoretical yield = 128 g; actual yield = 112 g
b. theoretical yield = 0.0567 g; actual yield = 0.0238 g
2. Calculate the actual yield based on the provided data
a. theoretical yield = 23.98 kg; percent yield = 78.4%
b. theoretical yield = 0.56 mg; percent yield = 2.34%
3. I react 0.42 g of nitrogen monoxide with excess oxygen in the lab and produce 0.53 g of nitrogen
dioxide. What is my percent yield for this reaction? (Hint: You have to write out the reaction to
solve.)
4. Al + I2  AlI3
a. What is the theoretical yield(g) if the reaction started with 1.20 mol Al ?
b. What is the theoretical yield (g) if the reaction started with 1.20 g Al?
c. What is the percent yield of 30 g of Al using the theor yield from a?
d. What is the percent yield of 30 g of Al using the theor yield from b?
5. Al2S3 + H2O  Al(OH)3 + H2S
a. If 15 g of aluminum sulfide completely react what is the theor yield of aluminum hydroxide?
b. What is the percent yield of 6.53 g of aluminum hydroxide?
6. Compound X is the product of a reaction. The reaction had a 75.2 % yield. Stoichiometric
calculations predicted the product to have a mass of 23.5 grams. What was the actual yield?
Adv Chem
11
Test Review: Stoichiometry
The following concepts and calculations are important to review:
 Make sure you can recognize balanced chemical equations
 Understand what the equation tells you about relationships between reactants, between products,
and between reactants and products
 Remember the relationships between moles and: atoms, formula units, molecules, molar mass,
formula weight, molecular weight, and volume of an ideal gas
 Be able to write conversion factors from the balanced chemical reaction
 Be able to compute the # grams, moles, molecules, atoms using the relationships between
reactants and products in a balanced chemical reaction
1. What is the difference between the theoretical yield and the actual yield of a reaction?
2. You will need to read this one a couple of times…There are x grams of carbon in a certain molecular
compound. After a chemical reaction, this carbon is found in two products. However, the total mass
of carbon in the two products is x grams. Does this observation contradict the Law of Conservation
of Mass? Explain.
3. Consider this reaction: 2 C6H6 + 15 O2  12 CO2 + 6 H2O
a. True or false: The mole ratio of C6H6 to CO2 is 6:1.
b. If 78 grams of C6H6 are reacted with excess O2, how many grams of water are produced?
c. If 10.00 grams of oxygen are reacted with C6H6, how many grams of oxygen will be found in
the two products?
Balance each equation and then answer the questions that follow each equation.
4. Si2H3 + O2  SiO2 + H2O
a. What is the mole ratio of silicon hydride to oxygen?
b. How many moles of water are produced from 5 moles of silicon hydride?
c. How many molecules of oxygen must be reacted to produce 5.62 x 1021 molecules of water?
5. Al(OH)3 + H2SO4  Al2(SO4)3 + H2O
a. How many grams of aluminum sulfate can be produced from 10 g of sulfuric acid?
b. How many moles of aluminum hydroxide are needed to react completely with 1 mole of
sulfuric acid?
6. Fe2(SO4)3 + KOH  K2SO4 + Fe(OH)3
a. 3.41 x 1026 formula units of iron (III) sulfate will produce how many moles of iron (III)
hydroxide?
b. What is the mole ratio of potassium hydroxide to iron(III) hydroxide?
c. If 25.00 g of iron (III) hydroxide are produced in this reaction, how many grams of potassium
sulfate are also produced?
7. C7H6O2 + O2 ---> CO2 + H2O
a. If C7H6O2 contains 10.00 grams of carbon, how many grams of carbon are in the carbon
dioxide product?
b. How many moles of oxygen are required to produce 30.00 g of water?
c. How many liters of oxygen are required to produce 30.00 g or water?
Adv Chem
12
More
of the Review on Next Page!
8. CO2(g) + Ca(OH)2 (aq)  CaCO
3 (s) + H2O
a. Ten grams of calcium carbonate were produced when carbon dioxide was added to lime water
(calcium hydroxide in solution). What volume of carbon dioxide was needed?
9. NH3 + H2O  NH4OH
a. What volume of ammonia is needed to add to water to produce 11 moles of ammonium
hydroxide?
10. CO2 + H2O  H2CO3
a. How many grams of carbonic acid (H2CO3) are produced when 55 liters of carbon dioxide are
dissolved into water?
11. B2O3 + Mg  MgO + B
a. How many atoms of boron can be obtained from 10.00 g of B2O3?
b. How many moles of magnesium are required to produce 400.0 g of boron?
Adv Chem
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