Question

Answer 1

Fe₃O₄+4H₂-to>3Fe+4H₂O

0,1-----0,4mol

n_Fe3O4=\(\dfrac{23,2}{232}\)=0,1 mol

=>V_H2=0,4.22,4=8,96l

Answer 2

\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\\ 4H_2+Fe_3O_4\rightarrow\left(t^o\right)3Fe+4H_2O\\ n_{H_2}=4.n_{oxit}=4.0,1=0,4\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\)