Дано:
\(V(\mathrm{H_2})=56\ \text{дм}^3\), \(V_m=22{,}4\ \text{дм}^3/\text{моль}\)
Найти: \(m(\mathrm{Mg})\), \(m(\mathrm{MgSO_4})\).
Решение:
\[
\mathrm{Mg+H_2SO_4\rightarrow MgSO_4+H_2}
\]
\[ n(\mathrm{H_2})=\frac{56}{22{,}4}=2{,}5\ \text{моль} \]
По уравнению \(n(\mathrm{Mg})=n(\mathrm{MgSO_4})=2{,}5\ \text{моль}\).
\[ m(\mathrm{Mg})=2{,}5\cdot24=60\ \text{г} \]
\[ M(\mathrm{MgSO_4})=24+32+4\cdot16=120\ \text{г/моль} \]
\[ m(\mathrm{MgSO_4})=2{,}5\cdot120=300\ \text{г} \]
Ответ: \(m(\mathrm{Mg})=60\ \text{г}\); \(m(\mathrm{MgSO_4})=300\ \text{г}\).
