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 \(N\)
\({n_{HN{O_3}}} = 3{n_{Fe{{(N{O_3})}_3}}} + 2{n_{Cu{{(N{O_3})}_2}}} + {n_{NO}}\)
\( = 0,05.3 + 0,15.2 + 0,15 = 0,6\)
\( \to {n_{HN{O_3}{\text{ tham gia}}}} = \frac{{0,6}}{{100\% - 20\% }} = 0,75{\text{ mol}}\)
\(\to {C_{M{\text{ HN}}{{\text{O}}_3}}} = \frac{{0,75}}{{0,5}} = 1,5M\)
