Bạn tham khảo nha!
a. `Fe_3O_4 + 4CO \overset{t^o}\to 3Fe + 2CO_2`
b. $n_{Fe}$ = $\dfrac{33,6}{56}$ = `0,6` `mol`
Theo PT, $n_{Fe_3O_4}$ = $\dfrac{1}{3}$. $n_{Fe}$ = $\dfrac{1}{3}$. `0,6` = `0,2` `mol`
=> $m_{Fe_3O_4}$ = `0,2 × 232` = `46,4` `g`
c. Theo PT, $n_{CO}$ = `0,8` `mol`
=> $V_{CO}$ = `0,8 × 22,4` = `17,92` `l`