Đáp án:
\(\begin{array}{l}
1)\\
\% {m_{Al}} = 17,42\% \\
\% {m_{Cu}} = 82,58\% \\
2)\\
a)\\
\% {m_{Cu}} = 25\% \\
\% {m_{Al}} = 75\% \\
b)\\
{C_M}HN{O_3} = 2M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Al}} = \dfrac{{0,15 \times 2}}{3} = 0,1\,mol\\
{n_{N{O_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{n_{Cu}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
\% {m_{Al}} = \dfrac{{0,1 \times 27}}{{0,1 \times 27 + 0,2 \times 64}} \times 100\% = 17,42\% \\
\% {m_{Cu}} = 100 - 17,42 = 82,58\% \\
2)\\
a)\\
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
{n_{N{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{Cu}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
\% {m_{Cu}} = \dfrac{{0,1 \times 64}}{{25,6}} \times 100\% = 25\% \\
\% {m_{Al}} = 100 - 25 = 75\% \\
b)\\
{n_{HN{O_3}}} = 0,2 \times 2 = 0,4\,mol\\
{C_M}HN{O_3} = \dfrac{{0,4}}{{0,2}} = 2M
\end{array}\)
