On passing 0.1 Faraday of electricity through AlCl3 the amount of aluminium metal deposited on the cathode is (at mass of Al = 27)0.9 g how?
回答 (1)
2016-05-31 5:33 am
✔ 最佳答案
Al³⁺ + 3e⁻ → Al1 mole of e⁻ carry 1 F of electricity.
No. of moles of e⁻ = 0.1 mol
According to the equation, mole ratio e⁻ : Al = 3 : 1
No. of moles of Al deposited = (0.1 mol) × (1/3) = 1/30 mol
Molar mass of Al = 27 g/mol
Mass of Al deposited = (27 g/mol) × (1/30 mol) = 0.9 g
收錄日期: 2021-04-18 14:54:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160530210919AA1TRFi