How much gas will be released when 16.2 grams of aluminum interacts with hydrochloric acid?

univerkov | November 24, 2020 | education

2Al + 6HCl = 2AlCl3 + 3H2 ↑
1) n (Al) = m / Mr = 16.2 / 27 = 0.6 (gram / mol)
2) n (Al) / n (H2) = 2/3 = 0.6 / x = 0.9 (gram / mol)
3) V (H2) = Vm * n = 22.4 * 0.9 = 20.16 (liters)
Answer: 20.16 liters

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