1. In the industrial process for the manufacture of sulphuric acid.
Explain why the escape of sulphur dioxide from the production plant is undesirable. ( 1 mark)
2. 10.0 g of copper(II) carbonate is added to 50.0 cm^3 of 1.00 M sulphuric acid. At the end if the reaction 1150 cm^3 of a certain gas are collected at room temperature and pressure.
(a) Calculate the theoretical mass of copper(II) sulphate crystal (CuSO4‧5H2O) that can be obtained. ( 2 marks)
Molar volume calculations的化學問題
2009-09-26 6:26 am
回答 (2)
2009-09-26 7:03 am
✔ 最佳答案
(1) Because sulphur dioxide is toxic and has choking smell when inhaled by human.(2a) According to the equation:
CaCO3 + H2SO4 --> CaSO4 + CO2 + H2O
when 1 mole of CO2 is liberated, 1 mole of CaSO4 will be formed.
Then,
No. of moles of CO2 formed = 1150/24000 = 0.04792
Hence 0.04792 moles of CuSO4 will also be formed and hence the theoretical mass will be:
0.04792 x (63.5 + 32.1 + 4 x 16 + 5 x 18) = 11.96 g
參考: Myself
2009-09-26 12:11 pm
1.
Sulphur dioxide is toxic.
OR: Sulphur dioxide can cause acid rain.
OR: Sulphur dioxide can shoot up the rate of asthma.
2.(a)
CuCO3 + H2SO4 → CuSO4 + H2O + CO2
CuSO4 + 5H2O → CuSO4•5H2O
Mole ratio CuCO3 : CuSO4•5H2O = 1 : 1
Molar mass of CuCO3 = 63.5 + 12 + 16x3 = 123.5 g/mol
No. of moles of CuCO3 formed = 10/123.5 = 0.08097 mol
Theoretical no. of moles of CuSO4•5H2O formed = 0.08097 mol
No. of moles of CuSO4•5H2O = 63.5 + 32.1 + 16x4 + 5x(1x2 + 16) = 249.6 g/mol
Theoretical mass of CuSO4•5H2O = 0.08097 x 249.6 = 20.21 g ... (Ans)
Sulphur dioxide is toxic.
OR: Sulphur dioxide can cause acid rain.
OR: Sulphur dioxide can shoot up the rate of asthma.
2.(a)
CuCO3 + H2SO4 → CuSO4 + H2O + CO2
CuSO4 + 5H2O → CuSO4•5H2O
Mole ratio CuCO3 : CuSO4•5H2O = 1 : 1
Molar mass of CuCO3 = 63.5 + 12 + 16x3 = 123.5 g/mol
No. of moles of CuCO3 formed = 10/123.5 = 0.08097 mol
Theoretical no. of moles of CuSO4•5H2O formed = 0.08097 mol
No. of moles of CuSO4•5H2O = 63.5 + 32.1 + 16x4 + 5x(1x2 + 16) = 249.6 g/mol
Theoretical mass of CuSO4•5H2O = 0.08097 x 249.6 = 20.21 g ... (Ans)
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