Đáp án:
\(\begin{array}{l}
a)\\
{m_{Al}} = 2,7g\\
{m_{Fe}} = 5,6g\\
b)\\
{C_M}{H_2}S{O_4} = 2,04M\\
c)\\
m = 37,1g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 8,3\\
1,5a + b = 0,25
\end{array} \right.\\
\Rightarrow a = b = 0,1\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
{m_{Fe}} = 8,3 - 2,7 = 5,6g\\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,25\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,25 \times 98}}{{10,91\% }} = 224,56g\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{224,56}}{{1,83}} = 122,71\,ml\\
{C_M}{H_2}S{O_4} = \dfrac{{0,25}}{{0,12271}} = 2,04M\\
c)\\
2Al + 6{H_2}S{O_4} \to 2A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2Fe + 6{H_2}S{O_4} \to 2F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
m = 0,05 \times 342 + 0,05 \times 400 = 37,1g
\end{array}\)
