дано
m(CH3COOH) = 300 g
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V(H2) - ?
2CH3COOH+Zn-->(CH3COO)2+H2
M(CH3COOH) = 60 g/mol
n(CH3COOH) = m(CH3COOH) / M(CH3COOH) = 300 / 60 = 5 mol
2n(CH3COOH) = n(H2)
n(H2) = 5*1 / 2 = 2. 5 mol
V(H2) = n(H2) * Vm = 2. 5 * 22. 4 = 56 L
56 л