Đáp án:
\( V = 2,24{\text{ lít}}\)
Giải thích các bước giải:
Ta có:
\({n_{Al}} = \frac{{13,5}}{{27}} = 0,5{\text{ mol = }}{{\text{n}}_{Al{{(N{O_3})}_3}}}\)
Ta có:
\({m_{Al{{(N{O_3})}_3}}} = 0,5.(27 + 62.3) = 106,5{\text{ gam}}\)
\( \to {m_{N{H_4}N{O_3}}} = 111,5 - 106,5 = 5{\text{ gam}}\)
\( \to {n_{N{H_4}N{O_3}}} = \frac{5}{{80}} = 0,0625{\text{ mol}}\)
Bảo toàn e:
\(3{n_{Al}} = 10{n_{{N_2}}} + 8{n_{N{H_4}N{O_3}}}\)
\( \to 0,5.3 = 10{n_{{N_2}}} + 0,0625.8\)
\( \to {n_{{N_2}}} = 0,1{\text{ mol}}\)
\( \to V = {V_{{N_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
