How many grams of salt will be obtained by the interaction of 40 g of aluminum and 40 g of chlorine.

univerkov | December 19, 2020 | education

Given: m (Al) = 40g m (Cl2) = 40g
Find: m (salt) -?
Decision. 2Al + 3Cl2 = 2AlCl3 n (Al) = m (Al) / M (Al) = 40/27 = 1.48 g / mol n (Cl2) = m (Cl2) / M (Cl2) = 40/71 = 0 , 56 g / mol 1.48 * 2 = 2.96 0.56 * 3 = 1.68 => Cl2 deficient, we are calculating it. n (AlCl3) = 0.56 * 2 = 1.12 mol m (AlCl3) = M (AlCl3) * n (AlCl3) = 133.5 * 1.12 = 134.62 g
Answer: 134.62g.

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