Bạn tham khảo:
$Fe_2O_3+3H_2SO_4 \to Fe_2(SO_4)_3+3H_2O$
$n_{Fe_2O_3}=0,025(mol)$
$n_{H_2SO_4}=0,075(mol)$
$a/$
$m_{ddH_2SO_4}=\frac{0,075.98.100}{9,8}=75(g)$
$b/$
$n_{Fe_2(SO_4)_3}=0,025(mol)$
$C\%_{Fe_2(SO_4)_3}=\frac{0,025.400}{4+75}.100\%=12,66\%$