Đáp án: Bên dưới.
Giải thích các bước giải:
- $n_{Cu}$ = $m_{Cu}$ / $M_{Cu}$ = $80/64= 1,25(mol)_{}$
$n_{Fe_2O_3}$ = $m_{Fe_2O_3}$ / $M_{Fe_2O_3}$ = $16/(2*56+3*16)=0,1(mol)_{}$
- $n_{O_2}$ = $m_{O_2}$ / $M_{O_2}$ = $48/(2*16)=1,5(mol)_{}$
$V_{O_2}$ = $n_{O_2}*22,4=1,5*22,4=33,6(l)$
$n_{H_2}$ = $m_{H_2}$ / $M_{H_2}$ = $4/(2*1)=2(mol)_{}$
$V_{H_2}$ = $n_{H_2}*22,4=2*22,4=44,8(l)$
$n_{CO_2}$ = $m_{CO_2}/$ $M_{CO_2}$ = $66/(12+2*16)=1,5mol)_{}$
$V_{CO_2}$ = $n_{CO_2}*22,4=1,5*22,4=33,6(l)$