Answer:
1. 0.338 moles of Fe
2. 0.700 moles of Fe
3. 48.3%
Explanation:
This is the reaction:
Fe₂O₃ + 3C → 2Fe + 3CO
We were told that we produce 18.9 g of Fe. Let's convert the mass to moles:
18.9 g . 1mol/ 55.85 g = 0.338 moles of Fe
Let's make a rule of three; ratio is 1:2.
1 mol of oxide can produce 2 moles of elemental iron
Then, 0.350 moles must produce (0.350 .2) / 1 = 0.700 moles of Fe
Let's determine the percent yield:
(Yield produced /Theoretical Yield) . 100 = 48.3 %
Answer:
The actual yield of iron in moles = 0.338 moles
The theoretical yield of iron in moles = 0.700 moles iron
The percentage yield = 48.3 %
Explanation:
Step 1: Data given
Number of moles Fe2O3 = 0.350 moles
carbon = in excess
Mass of Fe = 18.9 grams
Step 2: The balanced equation
Fe2O3 + C ⟶ 2 Fe + 3 CO
Step 3: Calculate moles Fe
For 1 mol Fe2O3 we need 1 mol C to produce 2 moles Fe and 3 moles CO
For 0.350 moles Fe2O3 we'll have 2*0.350 = 0.700 moles Fe
Step 4: Calculate mass Fe
Mass Fe = moles Fe * molar mass Fe
Mass Fe = 0.700 moles * 55.845 g/mol
Mass Fe = 39.10 grams = the theoretical yield
Step 5: Calculate % yield
% yield = (actual yield / theoretical yield) * 100%
% yield = (18.9 grams / 39.10 grams )*100%
% yield = 48.3 %
Step 6: Calculate moles Fe actual yield
Moles Fe = 18.9 grams / 55.845 g/mol
Moles Fe = 0.338 moles
% yield in moles = (0.338 moles / 0.700 moles ) *100 %
% yield in moles = 48.2 %
