Giải thích các bước giải:
\(\begin{array}{l}
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{S{O_2}}} = 0,25mol\\
\to {n_{Cu}} = {n_{S{O_2}}} = 0,25mol\\
\to {m_{Cu}} = 16g\\
\to {m_{CuO}} = 12g \to {n_{CuO}} = 0,15mol\\
a)\\
\% {m_{Cu}} = \dfrac{{16}}{{28}} \times 100\% = 57,14\% \\
\% {m_{CuO}} = \dfrac{{12}}{{28}} \times 100\% = 42,86\% \\
b)\\
{n_{{\rm{CuS}}{O_4}}} = {n_{Cu}} + {n_{CuO}} = 0,4mol\\
\to {m_{{\rm{CuS}}{O_4}}} = 64g\\
{n_{{H_2}S{O_4}}} = 2{n_{Cu}} + {n_{CuO}} = 0,65mol\\
\to {m_{{H_2}S{O_4}}} = 63,7g\\
\to {m_{{\rm{dd}}{{\rm{H}}_2}S{O_4}}} = \dfrac{{63,7}}{{83,3\% }} \times 100\% = 76,47g\\
\to {m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{S{O_2}}} = 88,47g\\
\to C{\% _{{\rm{CuS}}{O_4}}} = \dfrac{{64}}{{88,47}} \times 100\% = 72,34\%
\end{array}\)
