Mg + 2HCl = MgCl2 + H2
V(H2)=n*Vm=3 моль*22,4моль/л=67,2 л
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Mg + 2HCl = MgCl2 + H2
V(H2)=n*Vm=3 моль*22,4моль/л=67,2 л
Ответ:67,2 л