Đáp án:

\( \% {m_{Cu}} = 46,56\% \\\% {m_{CuO}} = 53,44\% \)

Giải thích các bước giải:

Phương trình hóa học

\(Cu + 2{H_2}S{O_4}\xrightarrow{{{t^o}}}CuS{O_4} + S{O_2} + 2{H_2}O\)

\(CuO + {H_2}S{O_4}\xrightarrow{{}}CuS{O_4} + {H_2}O\)

Giả sử có $100\,\, gam$ dung dịch \(H_2SO_4\)

\( \to {m_{{H_2}S{O_4}}} = 95,267\% .100 = 95,267{\text{ gam}}\)

\( \to {n_{{H_2}S{O_4}}} = \dfrac{{95,267}}{{98}} = 0,972{\text{ mol}}\)

Ta có:

\({n_{S{O_2}}} = {n_{Cu}} = a{\text{ mol}}\)

Bảo toàn khối lượng:

\({m_{dd}} = {m_{Cu}} + {m_{CuO}} + {m_{dd\;{{\text{H}}_2}S{O_4}}} - {m_{S{O_2}}}\)

\( = 64a + 80b + 100 - 64a = 80b + 100\)

\({n_{CuS{O_4}}} = a + b \to {m_{CuS{O_4}}} = 160(a + b)\)

\({n_{{H_2}S{O_4}{\text{ dư}}}} = 0,972 - (2a + b)\)

Vì \(C{\% _{CuS{O_4}}} = 4C{\% _{{H_2}S{O_4}}} \\\to {m_{CuS{O_4}}} = 4{m_{{H_2}S{O_4}}}\)

\( \to 160(a + b) = 4.98.(0,972 - 2a - b) \to 944a + 552b = 381,024\)

\(C{\% _{CuS{O_4}}} = \dfrac{{160(a + b)}}{{100 + 80b}} = 68,376\%\\ \to 160a + 109,2992b = 68,376\)

Giải được:

\(a=0,2626;b=0,2411\)

\( \to {m_{Cu}} = 16,8064{\text{ gam;}}{{\text{m}}_{CuO}} = 19,288{\text{ gam}}\)

\( \to \% {m_{Cu}} = \dfrac{{16,8064}}{{16,8064 + 19,288}} = 46,56\% \\\to \% {m_{CuO}} = 53,44\% \)