Đáp án:
1)
\(\begin{array}{l}
{V_{{H_2}}} = 19,04l\\
{m_{Mg}} = 9,6g\\
{m_{Al}} = 8,1g\\
{m_m} = 99,3g
\end{array}\)
2)
\(m = 78,5g\)
Giải thích các bước giải:
1)
\(\begin{array}{l}
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}S{O_4}}} = \dfrac{m}{M} = \dfrac{{83,3}}{{98}} = 0,85mol\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,85mol\\
{V_{{H_2}}} = n \times 22,4 = 0,85 \times 22,4 = 19,04l\\
hh:Mg(a\,mol),Al(b\,mol)\\
24a + 27b = 17,7(1)\\
a + \dfrac{3}{2}b = 0,85(2)\\
\text{Từ (1) và (2)} \Rightarrow a = 0,4;b = 0,3\\
{m_{Mg}} = n \times M = 0,4 \times 24 = 9,6g\\
{m_{Al}} = 17,7 - 9,6 = 8,1g\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,4mol\\
{m_{MgS{O_4}}} = n \times M = 0,4 \times 120 = 48g\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al}} = 0,15mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = n \times M = 0,15 \times 342 = 51,3g\\
{m_m} = {m_{A{l_2}{{(S{O_4})}_3}}} + {m_{MgS{O_4}}} = 48 + 51,3 = 99,3g
\end{array}\)
2)
\(\begin{array}{l}
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
hh:F{e_2}{O_3}(a\,mol),CuO(b\,mol)\\
6a + 2b = 1,4(1)\\
160a + 80b = 40(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,2;b = 0,1\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,4mol\\
{m_{FeC{l_3}}} = n \times M = 0,4 \times 162,5 = 65g\\
{n_{CuC{l_2}}} = {n_{Cu}} = 0,1mol\\
{m_{CuC{l_2}}} = n \times M = 0,1 \times 135 = 13,5g\\
m = {m_{FeC{l_3}}} + {m_{CuC{l_2}}} = 65 + 13,5 = 78,5g
\end{array}\)
