Đáp án:
\({V_{{O_2}}}= 3,36{\text{ lít}}\)
\( \% {m_{A{l_2}{O_3}}} = 79,07\% ;\% {m_{Al}} = 20,93\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)
BTKL:
\({m_{Al}} + {m_{{O_2}}} = {m_X}\)
\( \to 8,1 + {m_{{O_2}}} = 12,9{\text{ gam}}\)
\( \to {m_{{O_2}}} = 4,8{\text{ gam}} \to {{\text{n}}_{{O_2}}} = \frac{{4,8}}{{32}} = 0,15{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)
\({n_{A{l_2}{O_3}}} = \frac{2}{3}{n_{{O_2}}} = 0,1{\text{ mol}}\)
\(X\) chứa \(Al_2O_3\) và \(Al\) dư
\({m_{A{l_2}{O_3}}} = 0,1.(27.2 + 16.3) = 10,2{\text{ gam}}\)
\( \to \% {m_{A{l_2}{O_3}}} = \frac{{10,2}}{{12,9}} = 79,07\% \to \% {m_{Al}} = 20,93\% \)
