Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 36,48\% \\
\% {m_{Zn}} = 63,52\% \\
b)\\
m = 52,55g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 6HN{O_3} \to Fe{(N{O_3})_3} + 3N{O_2} + 3{H_2}O\\
Zn + 4HN{O_3} \to Zn{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
hh:Fe(a\,mol),Zn(b\,mol)\\
{n_{N{O_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
56a + 65b = 15,35\\
3a + 2b = 0,6\\
\Rightarrow a = 0,1;b = 0,15\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{15,35}} \times 100\% = 36,48\% \\
\% {m_{Zn}} = 100 - 36,48 = 63,52\% \\
b)\\
{n_{Fe{{(N{O_3})}_3}}} = {n_{Fe}} = 0,1\,mol\\
{n_{Zn{{(N{O_3})}_2}}} = {n_{Zn}} = 0,15\,mol\\
m = 0,1 \times 242 + 0,15 \times 189 = 52,55g
\end{array}\)