Đáp án:
a) \({\text{\% }}{{\text{m}}_{Fe}} = 41,17\% ;\% {m_{CuO}} = 58,83\% \)
b) \({{\text{m}}_{muối}} = 31,2{\text{ gam}}\)
c) \({{\text{m}}_{dd\;{{\text{H}}_2}S{O_4}}} = 200{\text{ gam}}\)
Giải thích các bước giải:
Gọi số mol Fe và CuO lần lượt là x, y.
\( \to 56x + 80y = 13,6\)
Phản ứng xảy ra:
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
\(CuO + {H_2}S{O_4}\xrightarrow{{}}CuS{O_4} + {H_2}O\)
Ta có:
\({n_{{H_2}}} = {n_{Fe}} = x = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}} \to {\text{y = 0}}{\text{,1 mol}}\)
\( \to {m_{Fe}} = 0,1.56 = 5,6{\text{ gam}} \to {\text{\% }}{{\text{m}}_{Fe}} = \frac{{5,6}}{{13,6}} = 41,17\% \to \% {m_{CuO}} = 58,83\% \)
Ta có:
\({n_{FeS{O_4}}} = {n_{Fe}} = 0,1{\text{ mol;}}{{\text{n}}_{CuS{O_4}}} = {n_{CuO}} = 0,1{\text{ mol}} \to {{\text{m}}_{muối}} = 0,1.(56 + 96) + 0,1.(64.96) = 31,2{\text{ gam}}\)
Ta có:
\({n_{{H_2}S{O_4}}} = {n_{Fe}} + {n_{CuO}} = 0,2{\text{ mol}} \to {{\text{m}}_{{H_2}S{O_4}}} = 0,2.98 = 19,6{\text{ gam}} \to {{\text{m}}_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{19,6}}{{9,8\% }} = 200{\text{ gam}}\)
