Question

giúp mk nhanh vs ah

Answer 1

a) \(n_{Fe_2O_3}=\dfrac{18}{160}=0,1125\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

          0,1125->0,3375

=> \(V_{H_2}=0,3375.22,4=7,56\left(l\right)\)

b) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1125}{1}>\dfrac{0,2}{3}\) => H₂ hết, Fe₂O₃ dư

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

                        0,2-------->\(\dfrac{0,4}{3}\)

=> \(m_{Fe}=\dfrac{0,4}{3}.56=\dfrac{112}{15}\left(g\right)\)