06-07 Formulae, Equations and the Mole, Set B Answer Key To print: Ctrl + P, paper A4, margins Default.
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06-07

Formulae, Equations and the Mole

Set BAnswer key
A

Multiple choice

9 marks
1

D60 dm3[1]

2 mol NaHCO3 give 1 mol CO2, so 5.0 mol give 2.5 mol. Volume = 2.5 × 24 = 60 dm3.

2

CRuO2[1]

Mass of Ru = 5.79 − 1.39 = 4.40 g. Moles: Ru = 4.40101 = 0.0436, O = 1.3916 = 0.0869. The ratio is 1 : 2.

3

C1.00 dm3 of 0.10 mol/dm3 potassium hydroxide, KOH[1]

A: 0.100 mol × 40 = 4.0 g. B: 0.025 mol × 98 = 2.45 g. C: 0.10 mol × 56 = 5.6 g. D: 0.02 mol × 36.5 = 0.73 g.

4

C45 cm3[1]

15 cm3 of N2 reacts with 45 cm3 of H2, so nitrogen is limiting and 15 cm3 of H2 is left. 30 cm3 of NH3 forms. Total = 15 + 30 = 45 cm3.

5

A0.5 mole of oxygen unreacted.[1]

3 mol of CO needs only 1.5 mol of O2, so 0.5 mol of O2 is left and 3 mol of CO2 forms.

6

C0.56 mol/dm3[1]

Ca(OH)2 + 2HCl → CaCl2 + 2H2O. Moles of Ca(OH)2 = 0.2 × 0.0264 = 0.00528 mol, so moles of HCl = 0.01056 mol. Concentration = 0.010560.0190 = 0.56 mol/dm3.

7

A0.050 mol/dm3[1]

Moles of K+ = 0.100 × 0.0200 = 0.00200 mol, now in 40.0 cm3 of solution. Concentration = 0.002000.0400 = 0.050 mol/dm3.

8

A25%[1]

Moles of CO2 = 3.024 = 0.125 mol = moles of Na2CO3. Mass = 0.125 × 106 = 13.25 g. Purity = 13.2553 × 100% = 25%.

9

C90.0%[1]

Moles of SO2 = 43.224 = 1.80 mol = moles of ZnS. Mass of ZnS = 1.80 × 97 = 174.6 g. Purity = 174.6194 × 100% = 90.0%.

B

Structured questions

20 marks
10
(a)

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) [2]

(b)

Volume ratio CH4 : O2 = 1 : 2. [1]

20 cm3 of CH4 needs 40 cm3 of O2, which is less than the 60 cm3 of O2 available. [1]

O2 is in excess, so CH4 is the limiting reagent. [1]

(c)

Volume of CO2 produced = 20.0 cm3 (ratio CH4 : CO2 = 1 : 1) [1]

Volume of O2 remaining = 60 − 40 = 20.0 cm3, so total = 20 + 20 = 40.0 cm3 [1]

(d)

0.7 × 20 cm3 = 14.0 cm3 [1]

11

2HNO3 + Mg → Mg(NO3)2 + H2. Moles of nitric acid = 0.030 × 0.80 = 0.024 mol, moles of Mg = 1024 = 0.41667 mol [1]

0.024 mol of nitric acid reacts with only 0.012 mol of Mg, so Mg is in excess and nitric acid is the limiting reagent. [1]

Moles of Mg(NO3)2 = 0.012 mol, so mass = 0.012 × 148 = 1.78 g [1]

12
(a)

NaH + H2O → NaOH + H2 [1]

(b)

Moles of NaH = 0.2424 = 0.01 mol and moles of H2O = 1.818 = 0.1 mol [1]

1 mol of NaH reacts with 1 mol of H2O, so NaH is the limiting reactant. [1]

Moles of NaOH formed = 0.01 mol [1]

13
(a)(i)

Mg(OH)2 + 2HCl → MgCl2 + 2H2O [1]

(ii)

Moles of HCl = 0.0276 × 0.220 = 0.006072 mol, so moles of Mg(OH)2 = 0.006072 ÷ 2 = 0.003036 mol [1]

Concentration = 0.0030360.0250 = 0.121 mol/dm3 [1]

(b)

0.12144 × (24 + 2 × 16 + 2 × 1) = 7.04 g [1]

(c)

1 dm3 of water dissolves 7.04 g, so volume = 10.07.0435 = 1.42 dm3 [1]