How many grams of aluminum reacted with hydrochloric acid if 89 g of aluminum chloride were obtained?

univerkov | December 3, 2020 | education

2Al + 2Cl3 = 2AlCl3
n (AlCl3) = m \ M = 89 \ 133.5 = 0.67 mol
n (Al) = n (AlCl3) = 0.67 mol
m (Al) = nM = 0.67 * 27 = 18.09 g
Answer: 18.09 g

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