Question

Answer 1

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: FeO + H₂ --t^o--> Fe + H₂O

__________0,1<------0,1

=> V_H2 = 0,1.22,4 = 2,24 (l)

Answer 2

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ PTHH:FeO+H_2\rightarrow Fe+H_2O\) 

Tỉ lệ PT:       1mol       :       1mol

Phản ứng: \(1mol\leftarrow1mol\)

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)