pthh: \(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
bđ__0,04____0,4
pư__0,04____0,04______0,04
kt__0_______0,36______0,04
\(m_{ddsaupư}=2,88+200=202,88g\)
\(C\%ddH_2SO_4=\dfrac{0,36.98.100}{202,88}\approx17,39\%\)
\(C\%ddFeSO_4=\dfrac{0,04.152.100}{202,88}\approx3\%\)