nFeO = 0,15 mol
FeO + 2HCl \(\rightarrow\) FeCl2 + H2O
0,15.....0,3..........0,15.......0,15
\(\Rightarrow\) mHCl = 0,3.36,5 = 10,95 (g)
\(\Rightarrow\) C%HCl = \(\dfrac{10,95.100\%}{100}\) = 10,95%
\(\Rightarrow\) C%muối sau phản ứng = \(\dfrac{19,05.100\%}{100}\) = 19,05%
\(\)