Answer:
1. theoretical yield of Fe = 105g.
2. Percentage yield of Fe = 83.71%
Explanation:
Step 1:
Data obtained from the question. This includes:
The equation for the reaction:
Fe2O3(s) + CO(g) -> Fe(s) + CO2(g)
Mass of Fe2O3 = 150g
Actual yield of Fe = 87.9g
Theoretical yield of Fe =?
Percentage yield of Fe =.?
Step 2:
Balancing the equation.
Fe2O3(s) + CO(g) -> Fe(s) + CO2(g)
The above equation can be balanced as follow:
There are 2 atoms of Fe on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Fe as shown below:
Fe2O3(s) + CO(g) -> 2Fe(s) + CO2(g)
There are a total of 4 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of CO and 3 in front of CO2 as shown below:
Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)
Now the equation is balanced.
Step 3:
Determination of the mass of Fe2O3 that reacted and the mass of Fe produced from the balanced equation. This is illustrated below:
Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)
Molar Mass of Fe2O3 = (56x2) + (16x3) = 112 + 48 = 160g/mol
Molar Mass of Fe = 56g/mol
Mass of Fe from the balanced equation = 2 x 56 = 112g
From the balanced equation,
160g of Fe2O3 reacted.
112g of Fe is produced.
Step 4:
Determination of the theoretical yield of Fe. This is illustrated below:
From the balanced equation,
160g of Fe2O3 produced 112g of Fe.
Therefore, 150g of Fe2O3 will produce = (150x112)/160 =105g of Fe.
Therefore, the theoretical yield of Fe is 105g.
Step 5:
Determination of the percentage yield of Fe. This is illustrated below:
Actual yield of Fe = 87.9g
Theoretical yield of Fe =105g
Percentage yield of Fe =.?
Percentage yield = Actual yield/Theoretical yield x100
Percentage yield = 87.9/105 x 100
Percentage yield = 83.71%
