Đáp án:
a. $Fe_2O_3+6HCl\to 2FeCl_3+3H_2O$
b. $n_{HCl}=\dfrac{73}{36,5}=2(mol)$
Theo PTHH: $n_{Fe_2O_3}=\dfrac{1}{6}.n_{HCl}=\dfrac 13(mol)$
$\to m_{Fe_2O_3}=160.\dfrac 13=53,33(g)$
$n_{FeCl_3}=\dfrac{1}{3}. n_{HCl}=\dfrac 23(mol)$
$\to m_{FeCl_3}=\dfrac 23.162,5=108,34(g)$
c. Áp dụng BTKL: $m_{Fe_2O_3}+m_{HCl}=m_{FeCl_3}+m_{H_2O}$
$→m_{H_2O}=53,33+73-108,34=17,99(g)$