Question

If 60g of rust (Fe2O₁) are made in the reaction below, how many Liters of oxygen must be used?

4Fe(s) + 30⬇️2(g) → 2Fe⬇️2O⬇️3

Answer 1

18.

Step-by-step explanation:

1) according to the reaction (M(Fe)=56; M(O₂)=32; V₀=22.4):

4Fe+3O₂⇒2Fe₂O₃; - ν(O₂)=0.75*ν(Fe);

2) ν(Fe)=m(Fe)/M(Fe)=60/56≈1.071 [mol];

3) ν(O₂)=0.75*1.071=0.804 [mol];

4) V(O₂)=V₀*ν(O₂); ⇒ V(O₂)=22.4*0.804=18 [lit].