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MODULE • Chemistry Form 4

EXERCISE / LATIHAN

3
–3
1 50 cm of 2 mol dm sulphuric acid is added to an excess of copper(II) oxide powder. Calculate the mass of copper(II)
sulphate formed in the reaction. [Relative atomic mass: H = 1, O = 16, Cu = 64, S = 32]
50 cm asid sulfurik 2 mol dm ditambah kepada serbuk kuprum(II) oksida berlebihan. Hitungkan jisim kuprum(II) sulfat yang terbentuk
3
–3
dalam tindak balas itu. [Jisim atom relatif: H = 1, O = 16, Cu = 64, S = 32]
M = 2 mol dm –3
V = 50 cm ? g
3
CuO(aq) + H SO (aq) CuSO (ak) + 2H O(l)
2 4 4 2
2 × 50
Number of moles of sulpuric acid = = 0.1 mol
1 000
From the equation, 1 mol CuO : 1 mol CuSO
4
0.1 mol CuO : 0.1 mol CuSO 4
Mass of CuSO = 0.1 mol × [64 + 32 + (16 × 4)] g mol = 16 g
–1
4
2 27.66 g of lead(II) iodide is precipitated when 2.0 mol dm of aqueous lead(II) nitrate solution is added to an excess
–3
of aqueous potassium iodide solution. Calculate the volume of aqueous lead(II) nitrate solution used. [Relative atomic
mass: I = 127, Pb = 207]
–3
27.66 g plumbum(II) iodida termendak apabila 2.0 mol dm larutan plumbum(II) nitrat akueus ditambahkan kepada larutan kalium iodida
akueus berlebihan. Hitungkan isi padu plumbum(II) nitrat yang digunakan. [Jisim atom relatif: I = 127, Pb = 207]
M = 2 mol dm
–3
V = ? cm 25 g
3
Pb(NO ) (aq) + 2KI(aq) PbI (s) + 2KNO (aq)
3 2 2 3
27.66
Mol of PbI = = 0.06 mol
2 (207 + 2 × 127)
From the equation, 1 mol PbI : 1 mol Pb(NO )
2 3 2
0.06 mol PbI : 0.06 mol Pb(NO )
2 3 2
n mol 0.06 mol
Volume of Pb(NO ) = M mol dm –3 = 2 mol dm –3 = 0.03 dm = 30 cm 3
3
3 2

3
–3
3 Zinc oxide powder is added to 100 cm of 2 mol dm nitric acid to form zinc nitrate. Calculate
Serbuk zink oksida ditambahkan kepada 100 cm asid nitrik 2 mol dm untuk membentuk zink nitrat. Hitungkan
–3
3
(i) the mass of zinc oxide that has reacted.
jisim zink oksida yang bertindak balas.
(ii) the mass of zinc nitrate produced. [Relative atomic mass: H = 1, O = 16, Cl = 35.5, Zn = 65]
jisim zink nitrat yang terhasil. [Jisim atom relatif: H = 1, O = 16, Cl = 35.5, Zn = 65]
(i) 2HNO (aq) + ZnO(s) Zn(NO ) (aq) + H O(l)
3 3 2 2
100 × 2
Number of moles of HNO = = 0.2 mol
3 1 000
From the equation, 2 mol of HNO : 1 mol of ZnO
3
0.2 mol of HNO : 0.1 mol of ZnO
3
Mass of ZnO = 0.1 × [65 + 16] = 8.1 g
(ii) From the equation, 2 mol of HNO : 1 mol of Zn(NO )
3 3 2
0.2 mol of HNO : 0.1 mol of Zn(NO )
3 3 2
–1
Mass of Zn(NO ) = 0.1 mol × [65 +[14 + (16 × 3)] × 2] g mol = 0.1 × 189 = 18.9 g
3 2

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