1)
Gọi số mol \(Al\) là \(x\) mol suy ra \(Fe\) là \(1,5x\) mol.
\( \to 27x + 56.1,5x = 22,2 \to x = 0,2\)
Đốt cháy hỗn hợp
\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)
\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)
Ta có:
\({n_{{O_2}}} = \frac{3}{4}{n_{Al}} + \frac{2}{3}{n_{Fe}} = \frac{3}{4}.x + \frac{2}{3}.1,5x = 0,35{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,35.22,4 = 7,84{\text{ lít}}\)
2)
Phản ứng xảy ra:
\(C + {O_2}\xrightarrow{{{t^o}}}C{O_2}\)
Ta có:
\({n_{C{O_2}}} = \frac{{17136}}{{22,4}} = 765{\text{ mol = }}{{\text{n}}_{C{\text{ lt}}}}\)
\( \to {m_{C{\text{ lt}}}} = 765.12 = 9180{\text{ g = 9}}{\text{,18 kg}}\)
\({m_C} = 12.90\% = 10,8{\text{ kg}}\)
Hiệu suất:
\(H = \frac{{9,18}}{{10,8}}.100\% = 85\% \)