Answer :
Given:
initial mass of Fe = 20.0 g
initial mass of oxygen gas = 11.53 g
mass of oxygen remained = 3.78 g
Solution:
The chemical reaction is expressed as:
4Fe + 3O2 --> 2Fe2O3
We first calculate the oxygen that is used in the reaction,
Mass oxygen used = 11.53 g - 3.78 g = 7.75 g O2
Moles O2 = 7.75 g O2 ( 1 mol / 32 g ) = 0.2422 moles O2
Moles Fe2O3 produced = 0.2422 moles O2 ( 2 mol Fe2O3 / 3 moles O2 ) = 0.1615 mol Fe2O3
Mass Fe2O3 produced = 0.1615 mol Fe2O3 ( 159.7 g / 1 mol ) = 25.79 g Fe2O3
Therefore, the mass of ferrous oxide produced would be 45.79 g Fe2O3.
initial mass of Fe = 20.0 g
initial mass of oxygen gas = 11.53 g
mass of oxygen remained = 3.78 g
Solution:
The chemical reaction is expressed as:
4Fe + 3O2 --> 2Fe2O3
We first calculate the oxygen that is used in the reaction,
Mass oxygen used = 11.53 g - 3.78 g = 7.75 g O2
Moles O2 = 7.75 g O2 ( 1 mol / 32 g ) = 0.2422 moles O2
Moles Fe2O3 produced = 0.2422 moles O2 ( 2 mol Fe2O3 / 3 moles O2 ) = 0.1615 mol Fe2O3
Mass Fe2O3 produced = 0.1615 mol Fe2O3 ( 159.7 g / 1 mol ) = 25.79 g Fe2O3
Therefore, the mass of ferrous oxide produced would be 45.79 g Fe2O3.