$a,PTHH:Cu+4HNO_3\to Cu(NO_3)_2+2NO_2\uparrow+2H_2O$

$Zn+4HNO_3\to Zn(NO_3)_2+2NO_2\uparrow+2H_2O$

$b,$ Theo PT và từ giả thiết ta có hệ:

$\begin{cases} 64.n_{Cu}+65.n_{Zn}=1,29\\ 2n_{Cu}+2n_{Zn}=n_{NO_2}=\dfrac{0,896}{22,4}=0,04 \end{cases}\\\Rightarrow n_{Zn}=n_{Cu}=0,01(mol)$

$\Rightarrow m_{Zn}=0,01.65=0,65(g)\\ \Rightarrow \%m_{Zn}=\dfrac{0,65}{1,29}.100\%=50,39\%\\ \Rightarrow \%m_{Cu}=100\%-50,39\%=49,61\%$

Đáp án:

\(\begin{array}{l}
\% {m_{Zn}} = 50,39\% \\
\% {m_{Cu}} = 49,61\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Zn + 4HN{O_3} \to Zn{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
b)\\
{n_{N{O_2}}} = \dfrac{{0,896}}{{22,4}} = 0,04\,mol\\
hh:Zn(a\,mol),Cu(b\,mol)\\
\left\{ \begin{array}{l}
65a + 64b = 1,29\\
2a + 2b = 0,04
\end{array} \right.\\
 \Rightarrow a = b = 0,01\,\\
\% {m_{Zn}} = \dfrac{{0,01 \times 65}}{{1,29}} \times 100\%  = 50,39\% \\
\% {m_{Cu}} = 100 - 50,39 = 49,61\% 
\end{array}\)