Theo đề bài ta có : nFeO3 = \(\dfrac{16}{160}=0,1\left(mol\right)\)
a) Ta có PTHH :
(1) \(Fe2O3+6HCl->2FeCl3+3H2O\)
0,1mol.............0,6mol.......0,2mol
\(\left(2\right)3NaOH+FeCl3->3NaCl+Fe\left(OH\right)3\downarrow\)
0,6mol...............0,2mol................................0,2mol
b) Ta có : mFe(OH)3 = 0,2.107 = 21,4(g)
c) Ta có :
VddNaOH = \(\dfrac{n}{CM}=\dfrac{0,6}{2}=0,3\left(l\right)\)
Vậy.............