Sửa đề : Dung dịch $Fe_3O_4 \to Fe_2(SO_4)_3$
$n_{Ba(OH)_2} = n_{Ba} = \dfrac{27,4}{137} = 0,2(mol)$
Fe2(SO4)3 + 3Ba(OH)2 → 2Fe(OH)3 + 3BaSO4
.........................0,2..................\(\dfrac{0,4}{3}\)............0,2.............(mol)
2Fe(OH)3 \(\xrightarrow{t^o}\) Fe2O3 + 3H2O
\(\dfrac{0,4}{3}\)....................\(\dfrac{0,2}{3}\)................(mol)
Suy ra:
\(x=\dfrac{0,2}{3}.160+0,2.233=57,27\left(gam\right)\)
Đúng 2
Bình luận (1)
