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$3Fe$ + $2O_{2}$ -> $Fe_{3}$$O_{4}$
a. $n_{Fe}$= $\dfrac{16,8}{56}$= $0,3$ mol
$n_{O2}$= $\dfrac{6,72}{22,4}$= $0,3$ mol
Theo tỉ lệ mol 2 chất, ta có:
$\dfrac{0,3}{3}$ < $\dfrac{0,3}{2}$
=> $O_{2}$ dư
=> $n_{O2 dư}$= $(0,15 - 0,1)$ =$0,05$ mol
b. $m_{Fe3O4}$= $0,1× (56× 3+ 16× 4)$= $23,2$ $g$