Question

cho FeO t/d vs 98g dd H2SO4 10%

a) m Feo

Answer 1

FeO +H2SO4--->FeSO4 +H2O

m\(_{H2SO4}=\frac{98.10}{100}=9,8\left(g\right)\)

n\(_{H2SO4}=\frac{9,8}{98}=0,1\left(mol\right)\)

Theo pthh

n\(_{FeO}=n_{H2SO4}=0,1mol\)

m\(_{FeO}=0,1.72=7,2\left(g\right)\)

Nhớ tích cho mình nhé

Answer 2

\(n_{H_2SO_4}=\frac{98.10}{98.100}=0,1\left(mol\right)\)

\(PTHH:FeO+H_2SO_4\rightarrow FeSO_4+H_2O\\ m_{FeO}=72.0,1=7,2\left(g\right)\)