дано
m(H2SO4) = 29.4 g
n(Fe)-?
Mg+H2SO4-->MgSO4+H2
M(H2SO4) = 98 g/mol
n(H2SO4) = m/M = 29.4 / 98 = 0.3 mol
V(H2) = n*Vm = 0.3 * 22.4 = 6.72 L
Fe2O3+3H2-->2Fe+3H2O
3n(H2) = 2n(Fe)
n(Fe) = 2*0.3 / 3 = 0.2 mol
ответ 0.2 моль
дано
m(H2SO4) = 29.4 g
n(Fe)-?
Mg+H2SO4-->MgSO4+H2
M(H2SO4) = 98 g/mol
n(H2SO4) = m/M = 29.4 / 98 = 0.3 mol
V(H2) = n*Vm = 0.3 * 22.4 = 6.72 L
Fe2O3+3H2-->2Fe+3H2O
3n(H2) = 2n(Fe)
n(Fe) = 2*0.3 / 3 = 0.2 mol
ответ 0.2 моль