11)
Bảo toàn e:
\(2{n_{Zn}} + {n_{Ag}} = {n_{N{O_2}}} \to {n_{N{O_2}}} = 0,1.2 + 0,2 = 0,4{\text{ mol}} \to {\text{V = 0}}{\text{,4}}{\text{.22}}{\text{,4 = 8}}{\text{,96 lít}}\)
Ta có:
\({n_{Zn{{(N{O_3})}_2}}} = {n_{Zn}} = 0,1{\text{ mol; }}{{\text{n}}_{AgN{O_3}}} = {n_{Ag}} = 0,2{\text{ mol}} \to {{\text{n}}_{HN{O_3}}} = 2{n_{Zn{{(N{O_3})}_2}}} + {n_{AgN{O_3}}} + {n_{N{O_2}}} = 0,1.2 + 0,2 + 0,4 = 0,8{\text{ mol}}\)
\({m_{HN{O_3}}} = 0,8.63 = 50,4{\text{ gam}} \to {{\text{m}}_{dd{\text{ HN}}{{\text{O}}_3}}} = \frac{{50,4}}{{12,7\% }} = 396,85{\text{ gam}}\)
12)
Gọi số mol Zn và Fe lần lượt là x, y.
\( \to 65x + 56y = 4,19{\text{ gam}}\)
Ta có: \({n_{NO}} = \frac{{1,344}}{{22,4}} = 0,06{\text{ mol}}\)
Bảo toàn e:
\(2{n_{Zn}} + 3{n_{Fe}} = 3{n_{NO}} \to 2x + 3y = 0,06.3\)
Giải được: x=0,03; y=0,04.
\( \to {m_{Zn}} = 0,03.65 = 1,95{\text{ gam}} \to {\text{\% }}{{\text{m}}_{Zn}} = \frac{{1,95}}{{4,19}} = 46,54\% \to \% {m_{Fe}} = 53,46\% \)
Ta có:
\({n_{Zn{{(N{O_3})}_2}}} = {n_{Zn}} = 0,03{\text{ mol; }}{{\text{n}}_{Fe{{(N{O_3})}_3}}} = {n_{Fe}} = 0,04{\text{ mol}} \to {{\text{n}}_{HN{O_3}}} = 2{n_{Zn{{(N{O_3})}_2}}} + 3{n_{Fe{{(N{O_3})}_3}}} + {n_{NO}} = 0,03.2 + 0,04.3 + 0,06 = 0,24{\text{ mol}} \to {{\text{C}}_{M{\text{ HN}}{{\text{O}}_3}}} = \frac{{0,24}}{2} = 0,12M\)
