Giải thích các bước giải:
$1. n_{Fe}=$ $\dfrac{5,6}{56}=0,1(mol)$
PTHH: $2Fe + 3Cl_2 → 2FeCl_3$
Theo PTHH $→n_{FeCl_3}=n_{Fe}=0,1(mol)$
$→m_{FeCl_3}=162,5*0,1=16,25(g)$
$2. $$n_{O_2}=$ $\dfrac{2,24}{22,4}=0,1(mol)$
$2KMnO_4 → K_2MnO_4 + MnO_2 + O_2$
Theo PTHH $→n_{KMnO_4}= 2n_{O_2}=2*0,1=0,2(mol)$
$→m= 0,2*158=31,6(g)$
$3. n_{O_2}= \dfrac{4,48}{22,4}=0,2(mol)$
$3Fe + 2O_2 → Fe_3O_4$
Theo PTHH $n_{Fe}=\dfrac{3}{2} n_{O_2}= \dfrac{3*0,2}{2}=0,3(mol)$
$→m=0,3*56=16,8(g)$