Find the volume of H2 that is released when 108 grams of aluminum reacts with sulfuric acid. What is the mass of the salt formed?

univerkov | December 10, 2020 | education

2Al + 3H2SO4 dil = Al2 (SO4) 3 + 3H2
n (Al) = 108/27 = 4 moles
n (H2) = 4: 2 * 3 = 6 mol
V (h2) = 6 * 22.4 = 134.4 liters
n (Al2 (SO4) 3) n = 4: 2 = 2 mol
m (Al2 (SO4) 3) = 2 * (27 * 2 + (32 + 16 * 4) * 3 = 2 * 342 = 684 grams

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