Page 34 - ACE YR IGCSE A TOP APPROACH TO CHEM

0.5 mol of Cr needs to react with 0.375 mol of O in a 3. Multiply the empirical formula by the number
2
3 calculated in step 2.
complete reaction. [0.5 mol × = 0.375 mol]. But there
4
is 0.25 mol of O available for the reaction only. So, 2 (a) 0.2 mol [1]
2
oxygen is the limiting agent. Number of moles of Cl = volume 3
The mole ratio of O to Cr O is 3 : 2. 2 24 dm
2 2 3 4.8 dm 3
2
Number of moles of Cr O = 0.25 × = 0.167 mol = 24 dm 3
2 3 3
Relative formula mass of Cr O = 152 = 0.2 mol
3
2
Mass of chromium(III) oxide = 152 × 0.167 mol (b) Relative formula mass of NaOCl = 74.5 [1]
= 25.384 g Number of moles of NaOCl = 0.2 mol [1]
= 25.4 g Mass = 14.9 g [1]
8 A Mass of NaOCl = number of moles × M r
NaOH + HCl ➞ NaCl + H O = 0.2 mol × 74.5
2
Number of moles of NaOH = 0.025 dm × 0.10 mol dm = 14.9 g
–3
3
= 0.0025 mol (c) Number of moles of NaOH = 2 × 0.2 mol
25 cm = 0.025 dm 3 = 0.4 mol [1]
3
The mole ratio of NaOH to NaCl is 1 : 1. number of moles
Relative formula mass of NaCl = 58.5 Volume of NaOH =
Mass of NaCl = 0.0025 mol × 58.5 concentration
= 0.146 g = 0.4 mol –3
2.0 mol dm
9 D = 0.2 dm 3 [1]
Relative formula mass of NaNO = 85
3
1.7 3 (a) 0.05 mol [1]
Number of moles of NaNO = 85 = 0.02 mol Relative formula mass of CaO = 56
3
NaNO ➞ Na + NO 3 – mass
+
3
There are 2 moles of ions formed when 1 mole of NaNO Number of moles of CaO = M
3
r
dissolves. 2.805
1 mole = Avogadro constant = 6.02 × 10 particles = 56
23
Number of particles of NaNO = 0.05 mol
3
23
= 0.02 × 2 × 6.02 × 10 particles
= 0.04 × 6.02 × 10 particles (b) 0.15 mol [1]
23
Number of moles of HCl = concentration × volume
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= 0.3 mol dm × 0.5 dm
–3
Li PO ➞ 3Li + PO 4 3– = 0.15 mol
+
4
3
3 moles of positive ions are produced.
Number of particles of Li PO = 3 × 6.02 × 10 particles (c) CaO [1]
23
3
4
24
= 1.806 × 10 particles 0.05 mol is 0.025 less than 0.075 mol. [1]
Ratio = CaO : HCl = 1 : 2
Part 2: Structured Questions If CaO is 0.05 mol, HCl should be 0.10 mol. But the
number of moles of HCl is 0.15 mol, it is 0.05 mol in
1 (a) Add up to be 100% [1] excess. CaO is the limiting reagent.
(b) C = 24.30 = 2.03 (d) 5.55 g [1]
12 Relative formula mass of CaCl = 111
H = 4.05 = 4.05 Mass of CaCl = 111 × 0.05 2
2
1 = 5.55 g
Cl = 71.65 = 2.02 [1] 4 (a) Pb(NO ) (aq) + 2KI(aq) ➞ PbI (s) + 2KNO (aq)
35.5 3 2 2 3
Ratio = C : H : Cl = 1 : 2 : 1 [1] Correct formula [1]

[1]
Balance the equation
Empirical formula: CH Cl [1]
2
Steps to calculate empirical formula: Correct state symbols [1]
1. Calculate the number of moles of each atom. Note: Must get the equation correctly, otherwise
2. Get the ratio by dividing the number of moles of cannot do the questions below.
each by the smallest number of moles. (b) (i) 0.02 mol [1]
3. Form the empirical formula. Number of moles of Pb(NO ) = 6.62
331
(c) 99 = 2 [1] 3 2 = 0.02 mol
49.5
Molecular formula = C H Cl [1] (ii) 0.05 mol 8.30 [1]
2
4
2
Steps to get molecular formula: Number of moles of KI = 116
1. Find the relative molecular mass of the molecule
based on empirical formula. = 0.05 mol
2. Divide the molecular formula by the mass (iii) KI / potassium iodide [1]
calculated from step 1. 0.05 mol is more than 0.04 mol. [1]
Cambridge IGCSE TM
146 Ace Your Chemistry
Answers.indd 146 3/4/22 3:54 PM