SCH3U-R.H.KING ACADEMY
Name:
UNIT 3: REVIEW
For all questions calculate molar masses to two decimal places and give answers with the correct number of
significant digits (remember: do not round your values until writing the final answer).
1.
Give the percentage composition for each compound: a) H2SO4, b) Ca(OH)2. [a) H = 2.06%, S = 32.69%, O = 65.25%, b) Ca
= 54.09%, O = 43.18%, H = 2.73%]
Molar Mass of H2SO4 = 2(2.02) + 32.06 + 4(16.00)
= 98.08 g/mol
H = 4.04
S=32.06
O=64.00
98.08
98.08
98.08
= 32.69% H
= 65.25% S
= 54.09% O
2.
Calculate the molar mass of a) H2SO4, b) Fe2(Cr2O7)3. [a) 98.08 g/mol, b) 759.70 g/mol]
a) 2H + 1S + 4O
=2(2.02) + 32.06 + 4(16.00)
= 98.08 g/mol
3.
Molar mass of Ca(OH)2= 40.08 + 2(16.00) +2(1.01)
= 74.10 g/mol
Ca=40.08
O=32.00
H=2.02
74.10
74.10
74.10
= 43.18% Ca
=2.73% O
= 2.06% H
b) 2Fe + 6Cr + 21O
= 2(55.84) + 6(52.00) + 21(16.00)
= 759.70 g/mol
a) How many moles are in 16 grams of CuCl2? b) How much does 70. moles of NaCl weigh? [a) 0.12 mol (= 16 g CuCl2 x 1
mol/134.45 g) b) 4100 g (= 70. mol NaCl x 58.44 g/mol)]
4.
a) # mol = 16g CuCl2 x 1 mol CuCl2
134.45g CuCl2
= 0.12 mol of CuCl2
b) # g = 70. mol NaCl x 58.44 g NaCl
1 mol NaCl
= 4100 g of NaCl
a) How many molecules are in exactly 4.00 moles of H2O? b) How many hydrogen atoms are in exactly 4.00 moles of
H2O? c) How many hydrogen atoms are in 0.173 moles of H2O? [ there are 6.02 x 1023 particles in a mole … a) 2.41 x 1024
23
24
23
23
23
molecules (4 x 6.02x10 );b) 4.82 x 10 atoms (4 x 2 x 6.02x10 );c) 2.08 x 10 atoms (0.173 x 2 x 6.02x10 )]
23
24
a) # molecules = 4.00 mol H2O x
6.02 x 10 molecules
1 mol H2O
23
b) # hydrogen atoms = 4.00 mol H2O x 6.02 x 10 molecules x
1 mol H2O
c) # H atoms = 0.173 mol H2O x
5.
= 2.41 x 10 molecules
24
2 H atoms
= 4.82 x 10 H atoms
1 molecule H2O
23
23
6.02 x 10 molecules x
1 mol H2O
2 H atoms
= 2.08 x 10 atoms
1 molecule H2O
What mass of magnesium oxide results when 56.3 g O2 combines with excess magnesium? [(2Mg + O2  2MgO) 142 g
MgO]
2Mg + O2 = 2MgO
#g Mg =
56.3g O2
x
1 mol O2
32.00g O2
x
2 mol MgO
1 mol O2
x
40.30g MgO
1 mol MgO
= 142 g MgO
R1
SCH3U-R.H.KING ACADEMY
6.
Label as simplest formula, molecular formula, or both: a) CuCl2, b) CO2, c) O2, d) C4H10.
a)
b)
c)
d)
7.
Name:
Simplest- not molecular, it is ionic
Simplest and molecular
Molecular
Molecular
A substance is 80% C and 20% hydrogen by mass. a) What is the simplest formula? b) What is the molecular formula
of the compound if the molar mass is 30 g/mol?
a) Assume we have 80g of substance- therefore 80g C and 20g H
# mol C = 80g C x 1 mol C
# mol H = 20g H x 1 mol H
6.7 mol of C, 20 mol of H
12.01 g C
1.01 g H
Divide by the smallest: 6.7/6.7=1, 20/6.7=3 The ratio is 1 C: 3 H. The simplest formula is CH 3.
b) molar mass of CH3 = 15.04 g/mol
molar mass of unknown substance = 30 g/mol
30/15.04=2 The multiplier is 2
Molecular formula C2H6
8.
Balance these equations: a) C40H82 + O2 → CO2 + H2O, b) H2O + Al4C3 → CH4 + Al(OH)3 [a) 2C40H82 + 121O2  80CO2 + 82H2O,
b) 12H2O + Al4C3  3CH4 + 4Al(OH)3]
a) 2C40H82 + 121O2  80CO2 + 82H2O
b) 12H2O + Al4C3  3CH4 + 4Al(OH)3
9. What four things may cause actual yields to differ from theoretical yields? [Not all product is recovered (e.g.
spattering),Reactant impurities (e.g. weigh out 100 g of chemical which has 20 g of junk),A side reaction occurs (e.g. MgO vs.
Mg3N2),The reaction does not go to completion
1. Not all product is recovered (e.g. spattering)
2. Reactant impurities (e.g. weigh out 100 g of chemical which has 20 g of junk)
3. A side reaction occurs (e.g. MgO vs. Mg 3 N 2 )
4. The reaction does not go to completion.
R2
SCH3U
Name:
Stoichiometry Review
Review and be able to work all problems from your Stoichiometry Notes, Stoichiometry Worksheet, and Limiting and
Excess Reactant Worksheet.
Complete the following problem:
1. Sodium and water react. How many grams of sodium will react with water to produce 4.00 moles of hydrogen? [184
grams]
2Na + 2H2O  2NaOH + H2
#g Na= 4.00 mol H
x
2 mol Na
1 mol H2
x
22.99 g Na
1 mol Na
= 184 g Na
2. How many moles of lithium chloride will be formed by the reaction of chlorine with 4.00 grams of lithium bromide?
-2
[0.0460 mol or 4.60X10 mol]
2LiBr + Cl2 = 2LiCl + Br2
# mol LiCl=
4.00g LiBr
x
1 mol LiBr
86.84 g LiBr
x
2 mol LiCl
2 mol LiBr
=0.0460 mol
3. Aluminum will react with sulfuric acid. a. How many moles of H2SO4 will react with 18 mol of aluminum? b. How
many moles of each product will form? [a. 27 mol; b. 27 mol hydrogen and 9.0 mol aluminum sulfate]
2Al + 3H2SO4  Al2(SO4)3 + 3H2
# mol H2SO4 = 18 mol Al
x
3 mol H2SO4
= 27 mol of H2SO4
2 mol Al
9 times the amount of each reactant and product in the balanced chemical equation will be formed – 27 mol hydrogen,
9.0 mol aluminum sulfate
4. What mass of acetylene, C2H2, will be produced from the reaction of 90. grams of calcium carbide with water in the
following reaction? [37 grams]
CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(s)
#g C2H2 =
90g CaC2
x
1 mol CaC2
64.10 g CaC2
x
1 mol C2H2
1 mol CaC2
x
26.04 g C2H2
1 mol C2H2
= 37 g of acetylene
R3
SCH3U
Name:
5. What mass of ZrCl4 can be produced if 862 grams of ZrSiO4 and 950. grams of chlorine are available to react according
to the following equation? [1.10X103 grams]
ZrSiO4 + 2Cl2  ZrCl4 + SiO2 + O2
# mol ZrSiO4 = 862 g ZrSiO4 x 1 mol
183.3 g
= 4.70 mol ZrSiO4
# mol Cl2 = 950 g x 1 mol
LF test: 4.70/1 = 4.70 ZrSiO4 is limiting
70.90 g
13.4/3= 6.70
= 13.4 mol Cl2
#g ZrCl4=
1 mol ZrCl4
1 mol ZrSiO4
4.70 mol ZrSiO4 x
x
x
233.02 g ZrCl4
1 mol ZrCl4
= 1100g
6. Heating zinc (II) sulfide in the presence of oxygen yields zinc (II) oxide and sulfur dioxide. If 1.72 moles of ZnS is
heated in the presence of 3.04 moles of O2, which reactant will be used up? [ZnS is the limiting reactant]
2ZnS + 3O2  2ZnO + 2SO2
Limiting factor test:
1.72/2 = 0.860 (for ZnS)
3.04/3 = 1.01 (for O2)
ZnS is the limiting reactant.
7. Aluminum and oxygen react. a. Which reactant is limiting if 0.32 moles of aluminum and 0.26 moles of oxygen are
available? b. How many moles of aluminum oxide are formed from the reaction of 0.00638 moles of oxygen and
0.00915 moles of aluminum? c. If 3.17 grams of aluminum and 2.55 grams of oxygen are available, which reactant is
limiting? [a. aluminum is limiting; b. 4.25 X 10-3 mol; c. oxygen is limiting]
4Al + 3O2  2Al2O3
a) 0.32/4 = 0.080 for aluminum, 0.26/3=0.086 for oxygen. Aluminum is the limiting reagent.
b) # mol Al2O3 =
0.00915 mol Al x
2 mol Al2O3
= 4.25 x 10-3 mol
4 mol Al
c) 3.17/26.98 = 0.117 mol Al
Limiting factor test:
0.117/4=0.029
2.55/32.00 = 0.080 mol O2
0.080/3= 0.026 – oxygen is the limiting reagent
R4
SCH3U
Name:
8. Bromine replaces iodine in magnesium iodide during a single replacement reaction. a. Which is the excess reactant
when 560 grams of MgI2 and 360 grams of Br2 react, and what mass remains? b. What mass of I2 is formed in the same
process? [a. Br2 is the excess and 39 grams will remain; b. 510 grams]
MgI2 + Br2  MgBr2 + I2
a) 560/278.1= 2.01 mol MgI2
360/159.8= 2.25 mol Br2
#g Br2 = 2.01 mol MgI2 x
- MgI2 is the limiting factor
1 mol Br2
1 mol MgI2
x
159.8g
1 mol Br2
= 321 g of Br2 are needed to react with 560 g of MgI2. 360-321=39g of Br2 remain.
# g I2 = 2.01 mol MgI2 x
1 mol I2
1 mol MgI2
x
x
253.80g
1 mol I2
= 510 g of iodine are produced.
9. If 8.87 grams of As2O3 is used in the below reaction and 5.33 gram of arsenic is produced, what is the present yield?
[79.3%]
2As2O3 + 3C  3CO2 + 4As
#g As = 8.87g As2O3
x
1 mol As2O3
x
197.84 g As2O3
4 mol As
2 mol As2O3
x
74.92 g
1 mol As
Theoretical yield = 6.72 g
Actual yield = 5.33 g
(5.33/6.72)x100=79.3% yield
10. Tungsten can be produced from its oxide by reacting the oxide with hydrogen at a high temperature according to
the following equation:
WO3 + 3H2  W + 3H2O
a. What is the percent yield if 56.9 grams of WO3 yields 41.4 grams of tungsten? b. How many moles of tungsten will be
produced from 3.72 grams of WO3 if the yield is 92.0%? [a. 91.8%; b. 0.0148 mol]
a) #g W =
56.9g WO3
x
1 mol WO3
231.84g
x
1 mol W
1 mol WO3
x
1 mol WO3
231.84
x
x
1 mol W
1 mol WO3
x
183.84g
1 mol W
Theoretical yield: 45.1 g
Actual yield: 41.4 g
(41.4/45.1) x 100 = 91.8 % yield
b) # mol W=
3.72g WO3
Theoretical yield: 0.0160 mol
92
100
=
x
0.0160
x = 0.0148 mol would be produced
R5