Đáp án:
\(\begin{array}{l}
b)\\
m = 18,2g\\
c)\\
\% {m_{CuO}} = 43,96\% \\
\% {m_{A{l_2}{O_3}}} = 56,04\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
b)\\
{m_{A{l_2}{O_3}}} = {m_{cr}} = 10,2g\\
{n_{Cu}} = \dfrac{m}{M} = \dfrac{{6,4}}{{64}} = 0,1\,mol\\
{n_{CuO}} = {n_{Cu}} = 0,1\,mol\\
m = {m_{CuO}} + {m_{A{l_2}{O_3}}} = 0,1 \times 80 + 10,2 = 18,2g\\
c)\\
\% {m_{CuO}} = \dfrac{8}{{18,2}} \times 100\% = 43,96\% \\
\% {m_{A{l_2}{O_3}}} = 100 - 43,96 = 56,04\%
\end{array}\)
