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Chemistry 1010
Fall (Loader) Version 1
Test # 2
[50 Marks Total]
Name: _____________________
October, 22nd 2004
MUN #: ______________________
Marking Scheme
QUESTIONS
Part A
Part B
VALUE
16
34
MARK
Total
Part A: Multiple choice questions. [Each Question = 2 marks]
1.
The density of table sugar is 0.849 g mL–1. What is the minimum volume of
the container required to store 2.5 kg of table sugar?
(a) 0.34 L
(b) 2.1 L
(c) 2.9 L
(d) 2.9 x 10–3 L
(e) 3.4 L
2.
Calcium nitrate and sodium phosphate solutions react as follows:
3 Ca(NO3)2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaNO3(aq)
If 0.33 moles of calcium nitrate react according to the equation, the number of
moles of sodium phosphate required for complete reaction is:
(a) 0.33 mol
(b) 0.22 mol
(c) 0.50 mol
(d) 0.66 mol
(e) 2.00 mol
3.
For the reaction
2 Fe(s) + 3 Cl2(g) → 2 FeCl3(s)
If 0.500 moles of iron are allowed to react with 0.600 moles of chlorine gas,
then the theoretical yield of iron(III) chloride is
(a) 0.300 mol
(b) 0.400 mol
(c) 0.500 mol
(d) 0.600 mol
(e) 1.10 mol
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CHEMISTRY 1010
4.
A solution was prepared by dissolving 45.0 g of sodium hydroxide, NaOH(aq)
and making the total volume up to 500.0 mL with water. The final
concentration of the sodium hydroxide solution was
(a) 0.444 mol L–1
(b) 0.563 mol L–1
(c) 1.78 mol L–1
(d) 2.25 mol L–1
(e) 22.5 mol L–1
5.
The mass of silver nitrate, AgNO3, (molar mass = 169.9 g mol–1) contained in
55.00 mL of a 0.150 mol L–1 solution of silver nitrate is
(a) 0.00825 g
(b) 1.14 g
(c) 1.40 g
(d) 11.4 g
(e) 121 g
6.
A test tube contains about 1 mL of 0.1 mol L–1 aqueous sodium chloride. A
student accidentally adds both 1 mL of 0.1 mol L–1 calcium hydroxide and
1 mL of 0.1 mol L–1 copper(II) nitrate to the sodium chloride solution and a
precipitate forms. The precipitate that forms is
(a) CaCl2(s)
(b) Cu(OH)2(s)
(c) NaOH(s)
(d) NaNO3(s)
(e) Ca(NO3)2(s)
7.
When calcium carbonate is strongly heated calcium oxide and carbon dioxide
gas are formed according to the reaction
CaCO3(s) heat
CaO(s) + CO2(g).
→
When 1.051 g of calcium carbonate (molar mass = 100.1 g mol–1) was heated
the remaining residue contained 0.460 g of calcium oxide, CaO(s), (molar
mass = 56.08 g mol–1). The calculated percentage yield of CaO(s) in the
reaction is
(a) 78.0 %
(b) 99.0%
(c) 43.8%
(d) 24.0%
(e) 56.2%
8.
What volume of 6.00 mol L–1 nitric acid must be measured and diluted with
water to give 250 mL of 0.500 mol L–1 nitric acid solution?
(a) 20.8 mL
(b) 48.0 mL
(c) 83.3 mL
(d) 3.00 mL
(e) 250 mL
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CHEMISTRY 1010
Part B
Answer all of the questions (34 Marks)
SHOW YOUR WORKING
B1.
(a)
The mineral fluorite is calcium fluoride, CaF2(s), and is used in the
preparation of hydrogen fluoride gas. The reaction is
CaF2(s) + H2SO4(l) → CaSO4(s) + 2 HF(g).
When 50.0 g of CaF2(s) (molar mass = 78.07 g mol–1), was treated
with 70.0 g of pure H2SO4(l) (molar mass = 98.08 g mol–1), 23.4 g
of hydrogen fluoride was obtained.
[5]
(i)
Calculate the theoretical yield in grams of HF for the
reaction.
50.0 g
moles of CaF2(s) = 78.07g mol −1 = 0.640 5 mol can give 2 x 0.6405 mol = 1.281 mol HF
moles of H2SO4(l) =
70.0 g
98.08 g mol −1
= 0.713 7 mol can give 2 x 0.7137 mol = 1.427 mol HF
Since the CaF2(s) can give the least HF it is the limiting reagent so the
theoretical yield in grams is = 1.281 mol HF x (1.0079 + 18.9984) g mol-1
= 25.63 g
Ans: 25.6 g
[2]
(ii)
Calculate the percentage yield of HF(g) obtained from the
experiment.
actual yield
23.4 g
The percentage yield = theoretical yield % 100% = 25.63 g % 100% = 91.3 %
Ans: 91.3%
(b)
Write net ionic equations for the reactions described.
(i)
a white precipitate forms when solutions of barium nitrate
and sodium sulfate are mixed.
Ba 2+ (aq) + SO 2−
4 (aq) d BaSO 4 (s)
[2]
(ii)
bubbles of carbon dioxide gas are produced when aqueous
hydrochloric acid is poured on to solid calcium carbonate.
2 H + (aq) + CaCO 3 (s) d Ca 2+ (aq) + CO 2 (g) + H 2 O(l)
[3]
(c)
If a sample of natural silver has atomic mass = 107.878 amu and is
a mixture of two isotopes:
isotope
silver-107
silver-109
isotopic mass/amu
106.905100
calculate
abundance/%
51.35
48.65
Calculate (show your working!) the isotopic mass of silver-109.
[5]
If we imagine 100 average atoms, the mass is made up of 51.35 of silver-107 and 48.65 of
silver-109 so the total mass of 100 =
(106.905100 amu x 51.35) + (calculate X 48.65) = 107.878 x 100 amu
so calculate = 108.90 amu
Ans: 108.9 amu
n
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CHEMISTRY 1010
B2.
For the following reaction for the reaction in acid solution
Mn 2+ (aq) + IO −4 (aq) d MnO 2 (s) + IO −3 (aq)
(a)
Write a balanced half-reaction for the OXIDATION half-reaction.
Mn 2+ (aq) + 2 H 2 O(l) d MnO 2 (s) + 4 H + (aq) + 2 e −
[3]
(b)
Write a balanced half-reaction for the REDUCTION half-reaction.
IO −4 (aq) + 2 H + (aq) + 2 e − d IO −3 (aq) + H 2 O(l)
[3]
(c)
Write a the balanced net ionic equation for the complete reaction.
Mn 2+ (aq) + H 2 O(l) + IO −4 (aq) d MnO 2 (s) + 2 H + (aq) + IO −3 (aq)
[3]
(d)
Rewrite the equation for the reaction done on basic solution.
Mn 2+ (aq) + IO −4 (aq) + 2 OH − (aq) d MnO 2 (s) + H 2 O(l) + IO −3 (aq)
[3]
[5]
B3.
The elemental analysis of a compound containing only chlorine, oxygen
and sulfur gives the following results:
Cl, 52.54 %; O, 23.71%; S, 23.76 %.
Calculate the empirical formula for the compound
Cl
O
S
52.54 g
23.71 g
23.76 g
mole in 100g
1.48195 mol
1.48193 mol
0.740995 mol
ratio
1.99996 mol
1.99992 mol
1
in 100 g
whole number
ratio
2
2
The empirical formula is Cl2O2S
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