Đáp án:
\({m_{Al}}= 71,469{\text{ kg}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2A{l_2}{O_3}\xrightarrow{{{t^o}}}4Al + 3{O_2}\)
Ta có:
\({n_{A{l_2}{O_3}}} = \frac{{150}}{{27.2 + 16.3}} = \frac{{25}}{{17}}{\text{ kmol}}\)
\( \to {n_{Al{\text{ lt}}}} = 2{n_{A{l_2}{O_3}}} = \frac{{25}}{{17}}.2 = \frac{{50}}{{17}}{\text{ kmol}}\)
\( \to {n_{Al}} = \frac{{50}}{{17}}.90\% = 2,647{\text{ kmol}}\)
\( \to {m_{Al}} = 2,647.27 = 71,469{\text{ kg}}\)