n(Mg) = 2 mol
V(H2) - ?Mg + 2HCl = MgCl2 + H2
за РХР: n(Mg) = n(H2) = 2 mol ;
V(H2) = 2 mol * 22.4 l / mol = 44.8 l
Ответ: 44.8 l .
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n(Mg) = 2 mol
V(H2) - ?
Mg + 2HCl = MgCl2 + H2
за РХР: n(Mg) = n(H2) = 2 mol ;
V(H2) = 2 mol * 22.4 l / mol = 44.8 l
Ответ: 44.8 l .