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MODULE • Chemistry Form 4

4 8.0 g of copper(II) oxide powder is added to excess dilute nitric acid and heated. Calculate the mass of copper(II) nitrate
produced. [Relative atomic mass: N = 14, O = 16, Cu = 64]
8.0 g serbuk kuprum(II) oksida dicampurkan kepada asid nitrik cair yang berlebihan dan dihangatkan. Hitungkan jisim kuprum(II) nitrat
yang terhasil. [Jisim atom relatif: N = 14, O = 16, Cu = 64]

CuO + 2HNO Cu(NO ) + H O
3 3 2 2
8 g
Mol of CuO = = 0.1 mol
(64 + 16)g mol –1
From the equation,
1 mol of CuO : 1 mol of Cu(NO )
3 2
0.1 of CuO : 0.1 mol of Cu(NO )
3 2
Mass of Cu(NO ) = 0.1 mol × 188 g mol = 18.8 g
–1
3 2

5 1.3 g of zinc reacts with excess dilute sulphuric acid. The products are zinc sulphate and hydrogen. Calculate the
volume of hidrogen gas released at STP. [Relative atomic mass: Zn = 65, 1 mol of gas occupies 22.4 dm mol at STP]
3
–1
1.3 g zink bertindak balas dengan asid sulfurik cair yang berlebihan. Hasil tindak balas adalah zink sulfat dan hidrogen. Hitungkan isi padu
–1
3
hidrogen yang terbebas pada STP. [Jisim atom relatif: Zn = 65, isipadu molar gas = 22.4 dm mol pada STP]

Answer/Jawapan: 448 cm 3

6 0.46 g of sodium burns completely in chlorine gas at room conditions to produce sodium chloride. Calculate the
volume of chlorine gas that has reacted. [Relative atomic mass: Na = 23, Molar volume of gas = 24 dm mol at room
3
–1
conditions]
0.46 g natrium terbakar lengkap dalam gas klorin pada keadaan bilik menghasilkan natrium klorida. Hitungkan isi padu klorin yang
3
–1
diperlukan untuk bertindak balas lengkap. [Jisim atom relatif: Na = 23, isi padu molar gas = 24 dm mol pada keadaan bilik]

Answer/Jawapan: 0.24 dm 3

7 The equation shows the combustion of propane gas.
Persamaan menunjukkan pembakaran gas propana.
C H + 5O 3CO + 4H O
3 8 2 2 2
3
720 cm of propane gas (C H ) at room conditions burns in excess oxygen. Calculate the mass of carbon dioxide formed.
3
8
[Relative atomic mass: C = 12, O = 16, Molar volume of gas = 22.4 dm mol at room conditions]
–1
3
720 cm gas propana (C H ) pada keadaan bilik terbakar dalam oksigen berlebihan. Hitungkan jisim karbon dioksida yang terbentuk.
3
3
8
[Jisim atom relatif: C = 12, O = 16, isi padu molar gas = 24 dm mol pada keadaan bilik]
–1
3

Answer/Jawapan: 3.96 g

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