1)

Gọi số mol \(Al;Mg\) lần lượt là \(x;y\)

\( \to 27x + 24y = 6,3\)

Ta có:

\({n_{N{O_2}}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol}}\)

Bảo toàn e:

\(3{n_{Al}} + 2{n_{Mg}} = {n_{N{O_2}}}\)

\( \to 3x + 2y = 0,6\)

Giải được: \(x=0,1;y=0,15\)

\( \to {m_{Al}} = 0,1.27 = 2,7{\text{ gam}}\)

\( \to \% {m_{Al}} = \frac{{2,7}}{{6,3}} = 42,86\%  \to \% {m_{Mg}} = 57,14\% \)

Ta có:

\({n_{Al{{(N{O_3})}_3}}} = {n_{Al}} = 0,1{\text{ mol}}\)

\({n_{Mg{{(N{O_3})}_2}}} = {n_{Mg}} = 0,15{\text{ mol}}\)

Bảo toàn \(N\):

\({n_{HN{O_3}{\text{ phản ứng}}}} = 3{n_{Al{{(N{O_3})}_3}}} + 2{n_{Mg{{(N{O_3})}_2}}} + {n_{N{O_2}}}\)

\( \to {n_{HN{O_3}}} = 0,1.3 + 0,15.2 + 0,6 = 1,2{\text{ mol}}\)

\( \to {n_{HN{O_3}{\text{ tham gia}}}} = \frac{{1,2}}{{100\%  - 25\% }} = 1,6\)

\( \to {C_{M{\text{ HN}}{{\text{O}}_3}}} = \frac{{1,6}}{{0,1}} = 16M\)

2)

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

\( \to 56x + 64y = 12,4\)

Ta có:

\({n_{NO}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}}\)

Bảo toàn e:

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

\( \to 3x + 2y = 0,15.3 = 0,45\)

Giải được:

\(x=0,05;y=0,15\)

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

\( \to \% {m_{Fe}} = \frac{{2,8}}{{12,4}} = 22,58\%  \to \% {m_{Cu}} = 77,42\% \)

Ta có:

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

\({n_{Cu{{(N{O_3})}_2}}} = {n_{Cu}} = 0,15{\text{ mol}}\)

Bảo toàn \(N\)

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

\( = 0,05.3 + 0,15.2 + 0,15 = 0,6\)

\( \to {n_{HN{O_3}{\text{ tham gia}}}} = \frac{{0,6}}{{100\%  - 20\% }} = 0,75{\text{ mol}}\)

\(\to {C_{M{\text{ HN}}{{\text{O}}_3}}} = \frac{{0,75}}{{0,5}} = 1,5M\)