Đáp án:
\(\begin{array}{l}
\% {m_{Cu}} = 78,05\% \\
\% {m_{Al}} = 21,95\% \\
m = 198g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
Al + 4HN{O_3} \to Al{(N{O_3})_3} + NO + 2{H_2}O\\
b)\\
hh:Cu(a\,mol),Al(b\,mol)\\
{n_{NO}} = \dfrac{{17,92}}{{22,4}} = 0,8\,mol\\
64a + 27b = 49,2\\
\dfrac{{2a}}{3} + b = 0,8\\
\Rightarrow a = 0,6;b = 0,4\\
\% {m_{Cu}} = \dfrac{{0,6 \times 64}}{{49,2}} \times 100\% = 78,05\% \\
\% {m_{Al}} = 100 - 78,05 = 21,95\% \\
{n_{Cu{{(N{O_3})}_2}}} = {n_{Cu}} = 0,6\,mol\\
{n_{Al{{(N{O_3})}_3}}} = {n_{Al}} = 0,4\,mol\\
m = 0,6 \times 188 + 0,4 \times 213 = 198g
\end{array}\)