Page 27 - Focus SPM KSSM F4 2020 - Chemistry

Chemistry Form 4 Chapter 3 The Mole Concept, Chemical Formula and Equation
SPM Highlights EXAMPLE 3.24

The following is a chemical equation. 2Na(s) + Cl (g) → 2NaCl(s)
2
What is the mass of sodium needed to react
K CO (aq) + CaCl (aq) → CaCO (s) + 2KCl(aq)
2 3 2 3 with 0.5 mole of chlorine?
Which of the following statements is true? [Relative atomic mass: Na, 23]
A The reactants are potassium carbonate solid and
calcium chloride solution. Solution
B Two moles of potassium carbonate react with one 2Na(s) + Cl (g) → 2NaCl(s)
mole of calcium chloride 2
C The products are calcium carbonate precipitate (2 mol) (1 mol)
and potassium chloride solution ? g 0.5 mol ①
D Two moles of calcium carbonate and one mole of Based on the equation, 1 mole of Cl reacts
potassium chloride are produced with 2 moles of Na. ② 2
Examiner’s tip So, 0.5 mole of Cl will react with (0.5 × 2)
2
Students need to examine each option carefully. Option mole of Na or 1 mole of Na. ③ Chapter
A is false because potassium carbonate used is in The mass of Na needed = number of moles
aqueous state and not in a solid state. Option B and D
are false because the coefficients in the equation show × molar mass ④
that 1 mole of potassium carbonate reacts with 1 mole of = 1 × 23 3
calcium chloride to produce 1 mole of calcium carbonate = 23 g
and 2 moles of potassium chloride.
Answer: C
EXAMPLE 3.25
2C H (g) + 9O (g) → 6CO (g) + 6H O(l)
6
2
2
3
2
Solving numerical stoichiometry What is the volume of carbon dioxide gas
problems produced at STP if 63 g of propene, C H is
6
3
1. Stoichiometry is a study of quantitative burnt completely?
composition of substances involved in chemical [Relative atomic mass: H, 1; C, 12. Molar volume
3
-1
reactions. : 22.4 dm mol at STP]
2. Based on a balanced chemical equation, we Solution
can solve various numerical problems. General 2C H (g) + 9O (g) → 6CO (g) + 6H O(l)
2
6
2
3
2
steps in solving problem: (2 mol) (6 mol)
3
63 g ? dm ①
① Gather information from the question. Number of moles of C H = Mass ÷ molar mass
3
6
Convert the given unit into the number = 63
of moles. 3(12) + 6(1)
= 63
42
② Compare the ratio of moles of the = 1.5 mol ①
related substances. Based on the equation, 2 moles of C H
6
3
produce 6 moles of CO .②
2
This means that 1 mole of C H produces 3
③ Calculate the answer in proportion of moles of CO . 3 6
2
moles. So, 1.5 moles of C H produces (1.5 × 3) moles
3
6
which are 4.5 moles of CO . ③
2
The volume of CO produced at STP
2
④ Convert the answer from moles to the = Number of moles × molar volume ④
required units. = 4.5 × 22.4 dm 3
= 100.8 dm 3
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03 SPM CHEMISTRY F4.indd 55 27/02/2020 11:23 AM