Đáp án:

\(C{\% _{FeS{O_4}}} = 3,65\% \)

Giải thích các bước giải:

Phản ứng xảy ra:

\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)

\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)

Gọi số mol \(Zn;Fe\) lần lượt là \(x;y\).

\( \to {n_{{H_2}}} = {n_{Zn}} + {n_{Fe}} = x + y = \frac{{22,4}}{{22,4}} = 1{\text{ mol}}\)

\({n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 1{\text{ mol}} \to {{\text{m}}_{{H_2}S{O_4}}} = 1.98 = 98{\text{ gam}}\)

\( \to {m_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{98}}{{10\% }} = 980{\text{ gam}}\)

BTKL:

\({m_X} + {m_{dd\;{{\text{H}}_2}S{O_4}}} = {m_Y} + {m_{{H_2}}}\)

\( \to 65x + 56y + 980 = {m_Y} + 1.2 \to {m_Y} = 65x + 56y + 978{\text{ gam}}\)

\({n_{ZnS{O_4}}} = {n_{Zn}} = x{\text{ mol}}\)

\( \to {m_{ZnS{O_4}}} = x.(65 + 32 + 16.4) = 161x\)

\( \to C{\% _{ZnS{O_4}}} = \frac{{161x}}{{65x + 56y + 978}} = 11,6022\% \)

Giải được: \(x=0,75;y=0,25\).

\( \to {m_{Zn}} = 0,75.65 = 48,75{\text{ gam}}\)

\({m_{Fe}} = 0,25.56 = 14{\text{ gam}}\)

Ta có:

\({n_{FeS{O_4}}} = {n_{Fe}} = 0,25{\text{ mol}}\)

\( \to {m_{FeS{O_4}}} = 0,25.(56 + 32 + 16.4) = 38{\text{ gam}}\)

\( \to C{\% _{FeS{O_4}}} = \frac{{38}}{{65x + 56y + 978}} = 3,65\% \)