Page 35 - ACE YR IGCSE A TOP APPROACH TO CHEM

If Pb(NO ) is 0.02 mol, KI should be 0.04 mol. (c) The amount that contains the same number
3 2
But the number of moles of Kl is 0.05 mol, of units as the number of carbon atoms in
it is 0.01 mol in excess. 12 grams of carbon-12. / 12 g of carbon-12 /
(iv) Relative formula mass of PbI = 334 [1] Avogadro constant / 6.02 × 10 particles /
23
2
6.68 g [1] atoms / molecules [1]
Mass of PbI = 0.02 × 334 = 6.68 g
2 (d) (i) 0.0005 mol [1]
5 (a) Blue / purple [1] Note: Must change cm to dm .
3
3
3
Mg(OH) is alkaline. 1 dm = 1000 cm 3
2
to green [1] Number of moles of K CO = 0.02 × 0.025
3
2
= 0.0005 mol

MgCl is a salt. Salt is neutral.
2 (ii) 0.08 mol dm [1]
–3
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(b) Number of moles of HCl = 0.025 × 3.5 Ratio = K CO : HCl = 1 : 2
3
2
= 0.088 mol [1] Number of moles of HCl = 2 × 0.0005
Note: Must change cm to dm . = 0.001 mol
3
3
1 dm = 1000 cm 3 0.001
3
Number of moles of Mg(OH) = 0.088 Concentration of HCl = 0.0125
2 2 –3
= 0.044 mol [1] = 0.08 mol dm
Ratio = HCl : Mg(OH) = 2 : 1 (e) 720 cm 3 [1]
2 Refer to the number of moles calculated for 7(b).
Concentration of Mg(OH) = 0.044 Ratio = K CO : CO = 1 : 1
2
2
2 0.0205 Volume of CO = 0.03 × 24 dm 3
3
= 2.15 mol dm [1] 2 = 0.72 dm = 720 cm 3
–3
3
(c) Mass of H O = 203 – 95 = 108 [1] 8 (a) C H (g) + 3.5O (g) ➞ 2CO (g) + 3H O(l)
2 2 6 2 2 2
x = 108 = 6 [1] Correct formula [1]
18 Balanced equation [1]
One mole of H O = 18 g
2 (b) Number of moles of C H = 0.2 mol [1]
6
2
6 (a) 34 g [1] Number of moles of CO = 0.4 mol [1]
2
From the information given, we can get the number Volume of CO = 9.6 dm [1]
3
2
of moles of copper. From the information given, calculate the number
6.4
Number of moles of Cu = of moles of ethane.
64 6
= 0.1 mol Number of moles of C H = 30
6
2
Ratio = Cu : AgNO = 1 : 2 = 0.2 mol
3
Number of moles of AgNO = 0.2 mol The gas produced is carbon dioxide.
3
Mass of AgNO = 0.2 mol × 170 Ratio = C H : CO = 1 : 2
3
6
2
2
= 34 g Number of moles of CO = 0.2 × 2 = 0.4 mol
2
(b) Mass of Ag = 0.2 × 108 = 21.6 g [1] Volume of CO = 0.4 × 24 = 9.6 dm 3
2
Percentage of silver formed = 96.3% [1] 9 (a) Relative atomic mass [1]
Ratio = Ag : AgNO = 1 : 1
3
Number of moles of Ag = 0.2 mol (b) Butane is a covalent compound. [1]
Mass of Ag = 0.2 × 108 = 21.6 g Potassium fluoride is an ionic compound. [1]
Formula to calculate the percentage of yield Relative molecular mass is for a covalent compound.
actual yield Relative formula mass is for an ionic compound.
= × 100%
calculated yield (c) (i) 0.0025 mol [1]
20.8 Note: Must change cm to dm .
3
3
= × 100%
21.6 1 dm = 1000 cm 3
3
= 96.3% Number of moles of Y = 0.06 = 0.0025 mol
24
7 (a) K CO (s) + 2HCl(aq) ➞ 2KCl(aq) + CO (g) + (ii) 71 [1]
3
2
2
H O(l) [1] Relative molecular mass of Y = 0.1775
2 0.0025
(b) Number of moles of K CO = 0.03 mol [1] = 71
2
3
4.15 (iii) Chlorine [1]
Number of moles of K CO =
2 3 138 Based on the information given, it is a gas
= 0.03 mol at room temperature. The mass number of
Number of moles of HCl = 0.06 mol [1] chlorine is 35.5. Chlorine is diatomic at room
Ratio = K CO : HCl = 1 : 2 temperature.
2
3
Number of moles of HCl = 2 × 0.03 mol 35.5 × 2 = 71
= 0.06 mol
Answers 147
Answers.indd 147 3/4/22 3:54 PM