Page 28 - Focus SPM KSSM F4 2020 - Chemistry

Chemistry Form 4 Chapter 3 The Mole Concept, Chemical Formula and Equation

EXAMPLE 3.26 Checkpoint 3.4
Q1 Write balanced chemical equations for the
2Al(s) + 3CuO(s) → Al O (s) + 3Cu(s) following reactions.
2
3
If 2 g of excess aluminium is reacted with 0.06 (a) Sulphur trioxide + water → sulphuric acid
mole of copper(II) oxide, calculate the mass of (b) Magnesium + carbon dioxide gas →
magnesium oxide powder + carbon powder
aluminium left behind. (c) Hydrogen gas reacts with nitrogen gas to
[Relative atomic mass: Al, 27] produce ammonia gas.
Solution (d) Silver nitrate solution reacts with metallic
2Al(s) + 3CuO(s) → Al O (s) + 3Cu(s) copper to produce silver precipitate and
3
2
(2 mol) (3 mol) copper(II) nitrate solution.
?g of 2 g 0.06 mol ① Q2 The following equation shows the decomposition
Based on the equation, 3 moles of CuO react of potassium chlorate(V) by heat.
with 2 moles of Al. ② KClO (s) → KCl(s) + O (g)
3
2
So, 0.06 mole of CuO reacts with (a) Balance the equation.
( 0.06 × 2) mol of Al or 0.004 mole of Al. ③ (b) Calculate the volume of oxygen produced at
3
room conditions from the decomposition of
Chapter
3 The mass of Al that reacted 24.5 g of potassium chlorate(V).
= number of moles × molar mass ④ [Relative atomic mass: O, 16; Cl, 35.5; K,
39. Molar volume : 24 dm mol at room
-1
3
= 0.04 × 27 conditions.]
= 1.08 g
So, the mass of Al left behind = 2 - 1.08 g Q3 Lead is extracted according to the following
= 0.92 g equation.
2PbO(s) + C(s) → 2Pb(s) + CO (g)
2
SPM Highlights (a) How many moles of lead is extracted from
0.5 mole of lead(II) oxide?
Equation for the reaction between potassium and (b) How many lead atoms are produced if 44.6 g
oxygen: of lead(II) oxide is heated with excess carbon
4K(s) + O (g) → 2K O(s) powder?
2
2
What is the maximum mass of potassium oxide that is [Relative atomic mass: O, 16; Pb, 207.
formed when 19.5 g potassium is burned completely Avogadro’s constant: 6.02 × 10 mol .]
-1
23
in excess oxygen? [Relative atomic mass: O, 16; K,
39] Q4 Sodium reacts with water as follows.
A 23.5 g C 70.5 g 2Na(s) + 2H O(l) → 2NaOH(aq) + H (g)
2
2
B 47.0 g D 94.0 g (a) What are the products of the reaction?
(b) How many moles of sodium will react with 5
Examiner’s tip moles of water?
The amount of product depends only on the amount (c) Calculate the mass of sodium needed to be
of potassium as the oxygen is in excess. 23
4K(s) + O (g) → 2K O(s) used to produce 3.01 × 10 H molecules?
2
2
2
4 mol 2 mol [Relative atomic mass: Na, 23. Avogadro’s
-1
23
19.5 g ? g ① constant: 6.02 × 10 mol .]
The number of moles of K involved
= mass ÷ molar mass
19.5
= 39 = 0.5 mol ②
Based on the equation, 4 moles of K produce 2
moles of K O. This means that 2 moles of K produce
2
1 mole of K O. So, 0.5 mol of K produces (0.5 ÷ 2)
2
mole of K O or 0.25 mole of K O.
2
2
The mass of K O = number of moles × molar mass
2
= 0.25 × [2(39) + 16] g
= 23.5 g
Answer: A
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03 SPM CHEMISTRY F4.indd 56 27/02/2020 11:23 AM