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CHEM F.4
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a iii) By the equation: CuO + H2 --> Cu + H2O each mole of copper (II) oxide can produce 1 mole of copper metal. No. of moles of CuO = 9.54/(63.5 + 16) = 0.12 So mass of Cu obtained = 0.12 x 63.5 = 7.584 g (b ii) From the equation, eeach mole of nitrogen gas can produce 2 moles of ammonia if reacted completely. Hence, no. of moles of nitrogen gas = 560/(2 x 14) = 20 So theorectical mass of ammonia = 2 x 20 x 17 = 680 g So percentage conversion = 102/680 x 100% = 15%
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CHEM F.4
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a) Hydrogen can reduce copper(II) oxide to copper. The experiment can be carried out using the set-up shown below. http://cg-love.com/image-09E1_4BDD8B8C.gifiii) What mass of copper would be obtained if 9.54 g of the oxide were consumed in the reaction?b) Hydrogen is also used to manufacture ammonia by... 顯示更多 a) Hydrogen can reduce copper(II) oxide to copper. The experiment can be carried out using the set-up shown below. http://cg-love.com/image-09E1_4BDD8B8C.gif iii) What mass of copper would be obtained if 9.54 g of the oxide were consumed in the reaction? b) Hydrogen is also used to manufacture ammonia by reacting with nitrogen according to the following equation: N2(g) + 3H2(g) ?2NH3(g) ii) In the manufacturing process, a sufficient amount of hydrogen is allowed to react with 560 g of nitrogen. 102 g of ammonia are obtained. What is the percentage conversion of nitrogen to ammonia? (Relative atomic masses: H = 1.0, O = 16.0, N = 14.0, Cu = 63.5)最佳解答:
a iii) By the equation: CuO + H2 --> Cu + H2O each mole of copper (II) oxide can produce 1 mole of copper metal. No. of moles of CuO = 9.54/(63.5 + 16) = 0.12 So mass of Cu obtained = 0.12 x 63.5 = 7.584 g (b ii) From the equation, eeach mole of nitrogen gas can produce 2 moles of ammonia if reacted completely. Hence, no. of moles of nitrogen gas = 560/(2 x 14) = 20 So theorectical mass of ammonia = 2 x 20 x 17 = 680 g So percentage conversion = 102/680 x 100% = 15%
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