Đáp án:
\(\begin{array}{l}
{m_{{\rm{dd}}{H_2}S{O_4}}} = 400g\\
\% Al = 49,09\% \\
\% Fe = 50,91\% \\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = 7,17\% \\
C{\% _{FeS{O_4}}} = 3,7\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
{n_{{H_2}S{O_{_4}}}} = {n_{{H_2}}} = 0,4mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,4 \times 98 = 39,2g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{39,2 \times 100}}{{9,8}} = 400g\\
hh:Al(a\,mol),Fe(b\,mol)\\
\frac{3}{2}a + b = 0,4(1)\\
27a + 56b = 11(2)\\
\Rightarrow a = 0,2;b = 0,1\\
{m_{Al}} = n \times M = 0,2 \times 27 = 5,4g\\
{m_{Fe}} = 11 - 5,4 = 5,6g\\
\% Al = \dfrac{{5,4}}{{11}} \times 100\% = 49,09\% \\
\% Fe = \dfrac{{5,6}}{{11}} \times 100\% = 50,91\% \\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,1mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = n \times M = 0,1 \times 294 = 29,4g\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,1mol\\
{m_{FeS{O_4}}} = n \times M = 0,1 \times 152 = 15,2g\\
{m_{ddspu}} = {m_{hh}} + {m_{dd{H_2}S{O_4}}} - {m_{{H_2}}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\, = 11 + 400 - 0,4 \times 2 = 410,2g\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{29,4}}{{410,2}} \times 100\% = 7,17\% \\
C{\% _{FeS{O_4}}} = \dfrac{{15,2}}{{410,2}} \times 100\% = 3,7\%
\end{array}\)
