Đáp án:
\(\% {m_{MgO}} = 40\% ;\% {m_{CuO}} = 60\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(MgO + {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + {H_2}\)
\(CuO + {H_2}S{O_4}\xrightarrow{{}}CuS{O_4} + {H_2}\)
Vì \(C{\% _{MgS{O_4}}} = C{\% _{CuS{O_4}}} \to {m_{MgS{O_4}}} = {m_{CuS{O_4}}} = x\)
\( \to {n_{MgS{O_4}}} = \frac{x}{{24 + 96}} = \frac{x}{{120}} = {n_{MgO}} \to {m_{MgO}} = \frac{x}{{120}}.40 = \frac{x}{3}gam\)
\({n_{CuS{O_4}}} = \frac{x}{{64 + 96}} = \frac{x}{{160}}{\text{ mol = }}{{\text{n}}_{CuO}} \to {m_{CuO}} = \frac{x}{{160}}.80 = \frac{x}{2}{\text{ gam}}\)
\( \to \% {m_{MgO}} = \frac{{\frac{x}{3}}}{{\frac{x}{3} + \frac{x}{2}}} = 40\% \to \% {m_{CuO}} = 60\% \)
