Quantitative Chemistry
1.
2.
3.
Assuming complete reaction, what volume of 0.200 mol dm–3 potassium hydroxide solution
(KOH(aq)), is required to neutralize 25.0 cm3 of 0.200 mol dm–3 aqueous sulfuric acid,
(H2SO4(aq))?
A.
12.5 cm3
B.
25.0 cm3
C.
50.0 cm3
D.
75.0 cm3
Which of the following compounds has/have the empirical formula CH2O?
I.
CH3COOH
II.
C6H12O6
III.
C12H22O11
A.
II only
B.
III only
C.
I and II only
D.
II and III only
Assuming complete reaction, what volume of 0.200 mol dm–3 HCl(aq) is required to neutralize
25.0 cm3 of 0.200 mol dm–3 Ba(OH)2(aq)?
A.
12.5 cm3
B.
25.0 cm3
C.
50.0 cm3
D.
75.0 cm3
Ô4. Under what conditions would one mole of methane gas, CH4, occupy the smallest volume?
5.
A.
273 K and 1.01×105 Pa
B.
273 K and 2.02×105 Pa
C.
546 K and 1.01×105 Pa
D.
546 K and 2.02×105 Pa
The complete oxidation of propane produces carbon dioxide and water as shown below.
C3H8 + __O2 __CO2 + __H2O
What is the total of the coefficients for the products in the balanced equation for 1 mole of
propane?
6.
A.
6
B.
7
C.
12
D.
13
What volume of 0.500 mol dm–3 HCl(aq) is required to react completely with 10.0 g of calcium
carbonate according to the equation below?
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
7.
8.
9.
A.
100 cm3
B.
200 cm3
C.
300 cm3
D.
400 cm3
Which sample has the least number of atoms?
A.
1 mol of H2SO4
B.
1 mol of CH3COOH
C.
2 mol of H2O2
D.
2 mol of NH3
The relative molecular mass (Mr) of a compound is 60. Which formulas are possible for this
compound?
I.
CH3CH2CH2NH2
II.
CH3CH2CH2OH
III.
CH3CH(OH)CH3
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
The empirical formula of a compound is C2H4O. Which molecular formulas are possible for this
compound?
I.
CH3COOH
10.
11.
II.
CH3CH2CH2COOH
III.
CH3COOCH2CH3
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
What is the total number of ions present in the formula, Al2(SO4)3?
A.
2
B.
3
C.
5
D.
6
Calcium carbonate decomposes on heating as shown below.
CaCO3  CaO + CO2
When 50 g of calcium carbonate are decomposed, 7 g of calcium oxide are formed. What is the
percentage yield of calcium oxide?
12.
A.
7
B.
25
C.
50
D.
75
What volume of 0.500 mol dm–3 sulfuric acid solution is required to react completely with 10.0
g of calcium carbonate according to the equation below?
CaCO3(s) + H2SO4(aq)  CaSO4(aq) + H2O(l) + CO2(g)
13.
A.
100 cm3
B.
200 cm3
C.
300 cm3
D.
400 cm3
The temperature in Kelvin of 1.0 dm3 of an ideal gas is doubled and its pressure is tripled. What
is the final volume of the gas in dm3?
A.
B.
C.
D.
14.
15.
A 4 g sample of sodium hydroxide, NaOH, is dissolved in water and made up to 500 cm3 of
aqueous solution. What is the concentration of the resulting solution?
A.
0.1 mol dm–3
B.
0.2 mol dm–3
C.
0.5 mol dm–3
D.
1.0 mol dm–3
Sodium reacts with water as follows.
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
1.15 g of sodium is allowed to react completely with water. The resulting solution is diluted to
250 cm3. Calculate the concentration, in mol dm–3, of the resulting sodium hydroxide solution.
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(Total 3 marks)
16.
The relative molecular mass of aluminium chloride is 267 and its composition by mass is 20.3%
Al and 79.7% chlorine. Determine the empirical and molecular formulas of aluminium chloride.
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(Total 4 marks)
17.
(i)
Calcium carbonate is added to separate solutions of hydrochloric acid and ethanoic acid
of the same concentration. State one similarity and one difference in the observations you
could make.
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(2)
(ii) Write an equation for the reaction between hydrochloric acid and calcium carbonate.
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(2)
(iii) Determine the volume of 1.50 mol dm–3 hydrochloric acid that would react with exactly
1.25 g of calcium carbonate.
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(3)
(iv)
Calculate the volume of carbon dioxide, measured at 273 K and 1.01×105 Pa, which
would be produced when 1.25 g of calcium carbonate reacts completely with the
hydrochloric acid.
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(2)
(Total 9 marks)
18.
(a)
Write an equation for the formation of zinc iodide from zinc and iodine.
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(1)
(b)
100.0 g of zinc is allowed to react with 100.0 g of iodine producing zinc iodide.
Calculate the amount (in moles) of zinc and iodine, and hence determine which
reactant is in excess.
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(3)
(c)
Calculate the mass of zinc iodide that will be produced.
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(1)
(Total 5 marks)
19.
An oxide of copper was reduced in a stream of hydrogen as shown below.
After heating, the stream of hydrogen gas was maintained until the apparatus had cooled.
The following results were obtained.
Mass of empty dish = 13.80 g Mass of dish and contents before heating = 21.75 g
dish and contents after heating and leaving to cool = 20.15 g
(a)
Mass of
Explain why the stream of hydrogen gas was maintained until the apparatus cooled.
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(1)
(b)
Calculate the empirical formula of the oxide of copper using the data above, assuming
complete reduction of the oxide.
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(3)
(c)
Write an equation for the reaction that occurred.
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(1)
(d) State two changes that would be observed inside the tube as it was heated.
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(2)
(Total 7 marks)
20.
Copper metal may be produced by the reaction of copper(I) oxide and copper(I) sulfide
according to the below equation.
2Cu2O + Cu2S  6Cu + SO2
A mixture of 10.0 kg of copper(I) oxide and 5.00 kg of copper(I) sulfide was heated until no
further reaction occurred.
(a)
Determine the limiting reagent in this reaction, showing your working.
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(3)
(b) Calculate the maximum mass of copper that could be obtained from these masses of
reactants.
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(2)
(Total 5 marks)
21.
State and explain what would happen to the pressure of a given mass of gas when its absolute
temperature and volume are both doubled.
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(Total 3 marks)
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(3)
(b)
Calculate the maximum mass of copper that could be obtained from these masses of
reactants.
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(2)
(Total 5 marks)
22.
(i)
Crocetin consists of the elements carbon, hydrogen and oxygen. Determine the empirical
formula of crocetin, if 1.00 g of crocetin forms 2.68 g of carbon dioxide and 0.657 g of
water when it undergoes complete combustion.
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(6)
(ii) Determine the molecular formula of crocetin given that 0.300 mole of crocetin has a mass
of 98.5 g
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(2)
(Total 8 marks)
23.
The reaction below represents the reduction of iron ore to produce iron.
2Fe2O3 + 3C  4Fe + 3CO2
A mixture of 30 kg of Fe2O3 and 5.0 kg of C was heated until no further reaction occurred.
Calculate the maximum mass of iron that can be obtained from these masses of reactants.
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(Total 5 marks)
Answers
1.
C
2.
C
3.
C
4.
5.
B
B
6.
7.
D
A
8.
C
9.
C
10.
C
11.
B
12.
B
13.
B
14.
B
15.
moles of Na = = 0.05;
moles of NaOH = 0.05;
Accept “same as moles of Na”
concentration = = 0.20 (mol dm–3)
Allow ECF from moles of NaOH
3
[3]
16.
AlCl or similar working (no penalty for use of 27 or 35.5);
empirical formula AlCl3;
molecular formula: n = = 2;
Al2Cl6;
Full credit can be obtained if the calculations are carried out
by another valid method. Two correct formulas but no valid
method scores [2 max].
[4]
17.
(i)
(ii)
bubbling/effervescence/dissolving of CaCO3/gas given off
produced); more vigorous reaction with HCl/OWTTE;
(do not accept CO2
2
2HCl(aq) + CaCO3(s)  CaCl2(aq) + CO2(g) + H2O(1);
[1] for correct formulas, [1] for balanced, state symbols not
essential.
2
(iii) amount of CaCO3 = (no penalty for use of 100); amount of HCl = 2×0.0125 = 0.0250
mol (allow ECF); volume of HCl = 0.0167 dm3/16.7 cm3 (allow ECF);
3
(iv) 1:1 ratio of CaCO3 to CO2 /use 0.0125 moles CO2 (allow ECF); (0.0125×22.4) = 0.28
dm3/280 cm3/2.8×10–4 m3 (allow ECF);
1
Accept calculation using pV=nRT.
[9]
18.
(a)
Zn + I2 → ZnI2;
Accept equilibrium sign.
(b)
(moles of) zinc = 1.530; (moles of) iodine = 0.3940;
ECF throughout. –1 (SF) possible.
(reacting ratio is 1:1, therefore) zinc is in excess;
above.
(c)
1
3
Must be consistent with calculation
(amount of zinc iodide = amount of iodine used = moles)
×(65.37 + 253.8) = 125.8 (g);
(mass of zinc iodide =
1
Use ECF throughout. –1 (SF) possible.
[5]
19.
(a)
to prevent (re)oxidation of the copper/OWTTE;
(b)
number of moles of oxygen = = 0.10; number of moles of copper = =
0.10; empirical formula = Cu (0.10) : O (0.10) = CuO;
Allow ECF.
Award [1] for CuO with no working.
1
3
Alternate solution
(c)
(d)
= 79.8%
= 20.2%
= 1.25
= 1.29
H2 + CuO → Cu + H2O;
Allow ECF.
1
(black copper oxide) solid turns red/brown; condensation/water vapour (on sides of test
tube);
2
Accept change colour. Do not accept reduction of sample
size.
[7]
20.
(a)
n(Cu2O) = 10.0×103÷143.1 = 69.9 mol;
n(Cu2S) = 5.00×103÷159.16 = 31.4 mol;
Penalise failure to convert kg  g once only.
Cu2S is the limiting reagent;
ECF from above answers.
(b)
n(Cu) = 6×n(Cu2S) = 6×31.4 = 188 mol;
m(Cu) = 188×63.55 = 11900 12000 g/11.9 12.0 kg;
If Cu2O given in (a), allow 3×n(Cu2O) and 3×n(Cu2O)×63.55.
3
2
Allow ECF from (a).
[5]
21.
overall there will be no change to the pressure;
double absolute temperature and the pressure doubles;
double volume and the pressure halves;
Apply ECF if points 2 and 3 are incorrect.
OR
Use PV = nRT, Since n and R are constant;
V and T are both doubled;
P will remain unchanged;
OR
OWTTE for mathematical interpretation
e.g. T  P, therefore 2P;
V  1/P, therefore ½P;
No change to P, ½P×2P = P;
3
[3]
22.
(i)
n(C)(= n(CO2) = 2.68 g÷44.01 g mol 1) = 0.0609 mol;
n(H)(= 2×n(H2O) = 0.657 g÷18.02 g mol 1) = 0.0729 mol;
m(C) = 0.0609 mol×12.01 g mol 1 = 0.731 g
and m(H) = 0.0729 mol×1.01 g mol 1 = 0.0736 g;
m(O) = (1.00 0.731 0.0736)g = 0.195g;
n(C)
0.0609
n(H)
0.0730
n(O)
0.195
0.0609
0.0609
0.0122
0.0730
0.0730
0.0122
16.00
0.0122
0.0122
0.0122
4.99
5.98
1.00;
empirical formula: C5H6O;
For C5H6 award [4 max].
6
Steps used to arrive at the correct amounts (in moles) are
required for full marks.
(ii)
M(crocetin) = 98.5 g÷0.300 mol = 328 (g mol 1);
molecular formula: C20H24O4;
ECF from (i).
2
[8]
23.
3
n(Fe2O3) = 30×10 ÷159.7/n(Fe2O3) = 188 mol;
n(C) = 5.0×103÷12.01/n(C) = 416 mol;
Fe2O3 is the limiting reagent or implicit in calculation;
n(Fe) = 2×n(Fe2O3) = 2×188 = 376 mol;
m(Fe) = 376×55.85 = 21 kg;
Accept 2 sig. fig. or 3 sig. fig., otherwise use
Correct final answers score [5].
Allow ECF.
1(SF).
[5]