a) \(n_{Fe_2O_3}=\dfrac{11,2}{160}=0,07\left(mol\right)\)
PTHH: Fe2O3 + 6HCl ---> 2FeCl3 + 3H2O
0,07---------------->0,14
=> mmuối = mFeCl3 = 0,14.162,5 = 22,75 (g)
b) mdd sau pư = 11,2 + 200 = 211,2 (g)
=> \(C\%_{FeCl_3}=\dfrac{22,75}{211,2}.100\%=10,77\%\)