Chemistry 1010 Sections 1, 2, 3 and 10 Loader (Fall 2004)
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Chemistry 1010
Handout 3: Moles and Stoichiometry
1.
2.
Calculate the empirical formula for the compounds with
analysis:
(a) 46.3% Li; 53.7% O
Ans: Li2O
(b) 68.4% Cr; 31.6% O
Ans: Cr2O3
(c) 22.6% K; 35.4% Cr; 38.0% O
Ans: K2CrO4
(d) 39.7% K; 27.9% Mn; 32.5% O
Ans: K2MnO4
Aluminum metal reacts with oxygen gas as follows:
4 Al(s) + 3 O2(g) → 2 Al2O3(s)
What mass of Al2O3(s) is produced when 3.5020 g of
aluminum is burned in air and what mass of oxygen gas is
used?
Ans: 6.169 g oxide and 3.1140 g oxygen used.
3.
Calculate the maximum mass of copper(II) oxide that
could be produced when 6.25 g of copper(II) carbonate is
heated: CuCO3(s) → CuO(s) + CO2(g)
Ans: 4.02 g
4.
What mass of CS2 is produced when 17.5 g of C are
reacted with 39.5 g of SO2 according to the equation:
5 C(s) + 2 SO2(g) → CS2(l) + 4 CO(g)?
Ans: C is the limiting reagent so the maximum
yield of CS2 is 22.1 g.
5.
When 2.16 g of osmium, Os, was converted to the oxide,
2.89 g of the oxide was obtained. What was the empirical
formula of the osmium oxide that formed?
Ans; OsO4
6.
When 0.15 mol of KO2 and 0.10 mol H2O are mixed the
following reaction takes place:
KO2 + H2O → KOH + O2
(a) Balance the equation.
Ans: 4 KO2 + 2 H2O → 4 KOH + 3 O2
(b) What is the limiting reagent?
Ans: KO2
(c) What is the theoretical yield of oxygen (in grams)?
Ans: 3.6 g of KO2
7.
Methyl salicylate (oil of wintergreen), C8H8O3, is prepared
by the reaction of salicylic acid, C7H6O3, with methanol,
CH4O.
C7H6O3 + CH4O → C8H8O3
When 1.50 g of salicylic acid, C7H6O3, reacts with 11.20 g
of methanol, CH4O, the reaction gave 1.24 g of methyl
salicylate, C8H8O3.
(a) Calculate the theoretical yield (in grams) of methyl
salicylate.
Ans: 1.65 g
(b) Calculate the percentage yield of methyl salicylate.
Ans: 75.0%
8.
Hydrogen gas can be made to react with nitrogen gas to
give ammonia gas:
____ H2(g) + ____ N2(g) → ____ NH3(g)
(a) Balance the above equation.
Ans: 3 H2(g) + N2(g) → 2 NH3(g)
(b) How many mole of ammonia can be made from 32.44
g of hydrogen gas?
Ans: 10.71 mol
(c) If 20.0 g of hydrogen is allowed to react with 20.0 g of
nitrogen, what is the theoretical (maximum possible)
yield of ammonia (in grams)?
Ans: 24.3 g
(d) If only 19.7 g of ammonia is obtained from the
reaction, what is the percentage yield?
Ans: 81.0%
9.
The reaction: SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(g)
produced 2.55 g of H2O when 12.2 g of SiO2 was treated
with a small excess of HF. (Note: Not all of the SiO2
reacts)
(a) What mass of SiF4 was formed?
Ans: 7.37 g
(b) What was the percentage yield of the SiF4?
Ans: 34.9%
(c) What mass of SiO 2 was left unreacted?
Ans: 7.95 g
10. When calcium carbonate, is heated strongly it
decomposes: CaCO3(s) → CaO(s) + CO2(g)
(a) If 49.98 g of CaCO3 is completely decomposed what
mass of CaO(s) is produced?
Ans: 28.00 g
(b) What mass of CaCO3 is needed to make 50.0 g of
carbon dioxide, CO2(g)?
Ans: 113.7 g
(c) A 250.0 g sample of CaCO3(s) was heated for 30
minutes and the residue was then cooled. The mass
of the cold residue was 178.8 g.
(i) Was the reaction complete? Explain.
Ans: No
(ii) If the reaction was incomplete, how much of each
of CaCO3 and CaO was present in the residue?
Ans: CaCO3 88.03 g and CaO 90.75 g
11. When 0.5000 g of aluminum, Al, reacted with chlorine
gas, Cl2(g), a chloride of aluminum was formed as pale
yellow solid that weighed 2.4710 g.
(a) What mass of chlorine has combined with the
aluminum?
Ans: 1.971 g
(b) What is the empirical formula for the compound?
Ans: AlCl3
12. Chromium metal can be produced in an electric furnace
by the reaction:
2 Cr2O3(s) + 3 Si(s) + 3 CaO(s) → 4 Cr(s) + 3 CaSiO3(s)
(a) How many mole of Cr2O3(s) are needed to make 5
mole of Cr(s).
Ans: 2½ mol
(b) How many mole of silicon, Si(s), are used in making 1
mole of chromium, Cr(s)?
Ans: 1.33
(c) How many tonne (metric tonne = 1000 kg) of
chromium can be made from 5.00 tonne of Cr2O3(s)?
Ans: 3.42 tonne
13. The concentration of the solution described:
(a) 100.00 mL of a solution containing 0.1500 mol of
NaNO3 (sodium nitrate).
Ans: 1.50 mol L-1
(b) 250.0 mL of a solution containing 5.5087 g of CaCl2
(calcium chloride).
Ans: 0.1985 mol L-1
(c) The concentration of the NaOH solution formed when
1.321 g of sodium reacted with water and the volume
of the final solution was 76.64 mL.
Ans: 0.7494 mol L-1
(d) 0.3010 g of CuSO4.5H2O(s) is dissolved and made up
to 200.00 mL with water.
Ans: 6.028 x 10-3 mol L-1
(e) The solution formed when 10.00 mL of a 0.500 mol.L-1
hydrochloric acid, HCl(aq), is diluted to 250.0 mL?
Ans: 0.0200 mol L-1
14. Calculate the number of moles and the mass of each
substance in the solutions.
(a) 20.00 mL of 0.150 mol.L-1 KNO3(aq)
Ans: 0.303 g
(b) 150 mL of 1.331 x 10-2 mol.L-1 magnesium chloride
solution.
Ans: 0.190 g