a)FeO +H2SO4-->FeSO4 +H2O
b) Ta có
m\(_{FeO}=\frac{36}{72}=0,5\left(mol\right)\)
m\(_{H2SO4}=\frac{100.15}{100}=15\left(g\right)\)
n\(_{H2SO4}=\frac{15}{98}=0,153\left(mol\right)\)
=>FeO dư
Theo PTHH
n\(_{FeSO4}=n_{FeO}=0,153\left(mol\right)\)
m\(_{FeSO4}=0,153.152=23,256\left(g\right)\)
c) C%=\(\frac{23,256}{100+3,6}.100\%=22,45\%\)