Đáp án:

\({m_{A{l_2}{O_3}}} = 15,3{\text{ gam}}\)

\({V_{hh{\text{ khí}}}} = 25,2{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2Al{(N{O_3})_3}\xrightarrow{{{t^o}}}A{l_2}{O_3} + 6N{O_2} + \frac{3}{2}{O_2}\)

Ta có:

\({n_{Al{{(N{O_3})}_3}}} = \frac{{63,9}}{{27 + 62.3}} = 0,3{\text{ mol}}\)

Ta có:

\({n_{A{l_2}{O_3}}} = \frac{1}{2}{n_{Al{{(N{O_3})}_3}}} = 0,15{\text{ mol}}\)

\( \to {m_{A{l_2}{O_3}}} = 0,15.(27.2 + 16.3) = 15,3{\text{ gam}}\)

\({n_{N{O_2}}} = 3{n_{Al{{(N{O_3})}_3}}} = 0,9{\text{ mol}}\)

\({n_{{O_2}}} = \frac{3}{4}{n_{Al{{(N{O_3})}_2}}} = 0,225{\text{ mol}}\)

\( \to {n_{hh{\text{khí}}}} = 0,9 + 0,225 = 1,125{\text{ mol}}\)

\( \to {V_{hh{\text{ khí}}}} = 1,125.22,4 = 25,2{\text{ lít}}\)