Đáp án:
\( \% {m_{Cu}} = 60\% ; \% {m_{CuO}} = 40\% \)
\({C_{M{\text{ CuS}}{{\text{O}}_4}{\text{ }}}} = 0,25M\)
\({C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = 1,75M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(CuO + {H_2}S{O_4}\xrightarrow{{}}CuS{O_4} + {H_2}O\)
Rắn không tan là \(Cu\)
\( \to {m_{Cu}} = 6{\text{ gam}} \to {{\text{m}}_{CuO}} = 10 - 6 = 4{\text{ gam}}\)
\( \to \% {m_{Cu}} = \frac{6}{{10}} = 60\% \to \% {m_{CuO}} = 40\% \)
\({n_{CuO}} = \frac{4}{{64 + 16}} = 0,05{\text{ mol = }}{{\text{n}}_{CuS{O_4}}}\)
\({n_{{H_2}S{O_4}}} = 0,2.2 = 0,4{\text{ mol}} \to {{\text{n}}_{{H_2}S{O_4}{\text{ dư}}}} = 0,4 - 0,05 = 0,35{\text{ mol}}\)
\( \to {C_{M{\text{ CuS}}{{\text{O}}_4}{\text{ }}}} = \frac{{0,05}}{{0,2}} = 0,25M\)
\({C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{0,35}}{{0,2}} = 1,75M\)