Đáp án:

a, 

\(\% {m_{Al}} = \dfrac{{0,1 \times 27}}{{35,2}} \times 100\%  = 7,67\% \)

\(\% {m_{Zn}} = \dfrac{{0,5 \times 65}}{{35,2}} \times 100\%  = 92,33\% \)

b,

\({m_{HN{O_3}}}{\rm{dd}} = \dfrac{{98,28 \times 100}}{{16}} = 614,25g\)

Giải thích các bước giải:

Áp dụng định luật bảo toàn electron, ta có:

\(\begin{array}{l}
Al \to A{l^{3 + }} + 3{\rm{e}}\\
Zn \to Z{n^{2 + }} + 2{\rm{e}}\\
2{N^{5 + }} + 10{\rm{e}} \to {N_2}\\
{n_{{N_2}}} = 0,13mol\\
 \to 3{n_{Al}} + 2{n_{Zn}} = 10{n_{{N_2}}} = 1,3
\end{array}\)

Mặt khác ta có: \(27{n_{Al}} + 65{n_{Zn}} = 35,2\)

Giải hệ phương trình, ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
27{n_{Al}} + 65{n_{Zn}} = 35,2\\
3{n_{Al}} + 2{n_{Zn}} = 1,3
\end{array} \right.\\
 \to {n_{Al}} = 0,1mol \to {n_{Zn}} = 0,5mol\\
 \to \% {m_{Al}} = \dfrac{{0,1 \times 27}}{{35,2}} \times 100\%  = 7,67\% \\
 \to \% {m_{Zn}} = \dfrac{{0,5 \times 65}}{{35,2}} \times 100\%  = 92,33\% 
\end{array}\)

\(\begin{array}{l}
{n_{HN{O_3}}} = 12{n_{{N_2}}} = 1,56mol\\
 \to {m_{HN{O_3}}} = 98,28g\\
 \to {m_{HN{O_3}}}{\rm{dd}} = \dfrac{{98,28 \times 100}}{{16}} = 614,25g
\end{array}\)

Bạn tham khảo:

$n_{Al}=a(mol$

$n_{Zn}=b(mol)$

$m_{hh}=27a+65b=35,2(1)$

$n_{N_2}=0,13(mol)$

Bt $e:$

$3.n_{Al}+2.n_{Zn}=0,13.10$

$ \to 3a+2b=1,3(2)$

$(1)(2)$

$a=0,1; b=0,5$

$a/$

$\%m_{Al}=\frac{0,1.27}{35,2}.100\%=7,67\%$

$\%m_{Zn}=92,33\%$

$m_{HNO_3}=\frac{0,26.63.100}{16}=102,375(g)$