Đáp án:
\( \% {m_{Al}}= 81,81\% ; \% {m_{Mg}} = 18,19\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)
\(2Mg + {O_2}\xrightarrow{{{t^o}}}2MgO\)
Ta có:
\({n_{Mg}} = \frac{{2,4}}{{24}} = 0,1{\text{ mol;}}{{\text{n}}_{{O_2}}} = \frac{{7,84}}{{22,4}} = 0,35{\text{ mol}}\)
Ta có:
\({n_{{O_2}}} = \frac{3}{4}{n_{Al}} + \frac{1}{2}{n_{Mg}} = 0,35\)
\( \to {n_{Al}} = 0,4{\text{ mol}}\)
\( \to {m_{Al}} = 0,4.27 = 10,8{\text{ gam}}\)
\( \to \% {m_{Al}} = \frac{{10,8}}{{10,8 + 2,4}} = 81,81\% \to \% {m_{Mg}} = 18,19\% \)