Đáp án:
\(\begin{array}{l}
a)\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 405g\\
b)\\
\% {m_{CuO}} = 80\% \\
\% {m_{A{l_2}{O_3}}} = 20\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
hh:CuO(a\,mol),A{l_2}{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
80a + 102b = 25,5\\
160a + 342b = 57,9
\end{array} \right.\\
\Rightarrow a = 0,255;b = 0,05\\
{n_{{H_2}S{O_4}}} = 0,255 + 0,05 \times 3 = 0,405\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,405 \times 98}}{{9,8\% }} = 405g\\
b)\\
\% {m_{CuO}} = \dfrac{{0,255 \times 80}}{{25,5}} \times 100\% = 80\% \\
\% {m_{A{l_2}{O_3}}} = 100 - 80 = 20\% \\
\end{array}\)
