Đáp án:
\(\begin{array}{l}
a)\\
M:Fe\\
b)\\
{C_\% }{H_2}S{O_4} = 9,8\% \\
c)\\
{V_{{H_2}}} = 4,48l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2M + n{H_2}S{O_4} \to {M_2}{(S{O_4})_n} + n{H_2}\\
{n_M} = 2{n_{{M_2}{{(S{O_4})}_n}}} \Leftrightarrow \dfrac{{11,2}}{{{M_M}}} = 2 \times \dfrac{{30,4}}{{2{M_M} + 96n}}\\
\Rightarrow {M_M} = 28n\,g/mol\\
n = 2 \Rightarrow {M_M} = 56g/mol \Rightarrow M:Fe\\
b)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2\,mol\\
{n_{{H_2}S{O_4}}} = {n_{Fe}} = 0,2\,mol\\
{C_\% }{H_2}S{O_4} = \dfrac{{0,2 \times 98}}{{200}} \times 100\% = 9,8\% \\
c)\\
{n_{{H_2}}} = {n_{Fe}} = 0,2\,mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l
\end{array}\)
