Đáp án:

\(\% {m_{Fe}} = 59,32\% ;\% {m_{Cu}} = 40,68\% \)

Giải thích các bước giải:

Gọi số mol \(Fe;Cu\) lần lượt là \(x;y\).

\( \to 56x + 64y = 4,72{\text{ gam}}\)

Ta có:

\({n_{NO}} = \frac{{1,568}}{{22,4}} = 0,07{\text{ mol}}\)

Bảo toàn e:

\(3{n_{Fe}} + 2{n_{Cu}} = 3{n_{NO}}\)

\( \to 3x + 2y = 0,07.3 = 0,21\)

Giải được: \(x=0,05;y=0,03\).

\( \to {m_{Fe}} = 0,05.56 = 2,8{\text{ gam}}\)

\( \to \% {m_{Fe}} = \frac{{2,8}}{{4,72}} = 59,32\%  \to \% {m_{Cu}} = 40,68\% \)

Ta có:

\({n_{Fe{{(N{O_3})}_3}}} = {n_{Fe}} = 0,05{\text{ mol}}\)

\({n_{Cu{{(N{O_3})}_2}}} = {n_{Cu}} = 0,03{\text{ mol}}\)

Bảo toàn nguyên tố \(N\).

\({n_{HN{O_3}}} = 3{n_{Fe{{(N{O_3})}_3}}} + 2{n_{Cu{{(N{O_3})}_2}}} + {n_{NO}}\)

\( = 0,05.3 + 0,03.2 + 0,07 = 0,28{\text{ mol}}\)

\( \to {m_{HN{O_3}}} = 0,28.63 = 17,64{\text{ gam}}\)

\( \to {m_{dd{\text{HN}}{{\text{O}}_3}}} = \frac{{17,64}}{{20\% }} = 88,2{\text{ gam}}\)

BTKL:

\({m_A} + {m_{dd{\text{HN}}{{\text{O}}_3}}} = {m_{dd{\text{ B}}}} + {m_{NO}}\)

\( \to 4,72 + 88,2 = {m_{dd\;{\text{B}}}} + 0,07.30 \to {m_{{\text{dd B}}}} = 90,82{\text{ gam}}\)

\({m_{Fe{{(N{O_3})}_3}}} = 0,05.(56 + 62.3) = 12,1{\text{ gam}}\)

\({m_{Cu{{(N{O_3})}_2}}} = 0,03.(64 + 62.2) = 5,64{\text{ gam}}\)

\( \to C{\% _{Fe{{(N{O_3})}_3}}} = \frac{{12,1}}{{90,82}} = 13,32\% \)

\( \to C{\% _{Cu{{(N{O_3})}_2}}} = \frac{{5,64}}{{90,82}} = 6,21\% \)