Vm-22,4 Моль-0,4
V=22,4•0,4=8,96л
Дано
m техн(Zn) = 6.5 g
W(прим)= 5%
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V(H2)-?
m чист(Zn) = 6.5 - (6.5 * 5% / 100%) = 6.175 g
Zn+H2SO4-->ZnSO4+H2
M(Zn)= 65 g/mol
n(Zn) = m/M = 6.175 / 65 = 0.095 mol
n(Zn) = n(H2) = 0.095 mol
V(H2) = n(H2)*Vm = 0.095*22.4 = 2.128 L , Vm = 22.4 L/mol
ответ 2.128 л