Câu 22:
$n_{Fe}= \frac{60}{56}= \frac{15}{14} mol$
$n_S= 0,9375 mol$
$Fe+ S \buildrel{{t^o}}\over\to FeS$
$n_{Fe dư}= \frac{15}{14}-0,9375= \frac{15}{112} mol= n_{H_2}$ (bảo toàn e)
$n_{FeS}= 0,9375 mol= n_{H_2S}$ (bảo toàn S)
$H_2S+ \frac{3}{2}O_2 \to SO_2+ H_2O$
$2H_2+ O_2 \to 2H_2O$
=> $n_{O_2}= 1,5n_{H_2S}+ 0,5n_{H_2}= \frac{165}{112}$
=> $V_{O_2}= 22,4.\frac{165}{112}= 33l$
