Đáp án:
\(\begin{array}{l}
{V_{{H_2}}} = 19,04l\\
{m_{Mg}} = 9,6g\\
{m_{Al}} = 8,1g\\
{m_m} = 99,3g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}S{O_4}}} = \dfrac{m}{M} = \dfrac{{83,3}}{{98}} = 0,85mol\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,85mol\\
{V_{{H_2}}} = n \times 22,4 = 0,85 \times 22,4 = 19,04l\\
hh:Mg(a\,mol),Al(b\,mol)\\
24a + 27b = 17,7(1)\\
a + \dfrac{3}{2}b = 0,85(2)\\
\text{Từ (1) và (2)} \Rightarrow a = 0,4;b = 0,3\\
{m_{Mg}} = n \times M = 0,4 \times 24 = 9,6g\\
{m_{Al}} = 17,7 - 9,6 = 8,1g\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,4mol\\
{m_{MgS{O_4}}} = n \times M = 0,4 \times 120 = 48g\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al}} = 0,15mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = n \times M = 0,15 \times 342 = 51,3g\\
{m_m} = {m_{A{l_2}{{(S{O_4})}_3}}} + {m_{MgS{O_4}}} = 48 + 51,3 = 99,3g
\end{array}\)
