Answer:
The answer to your question is 68.8%
Explanation:
Data
Iron = 3.8 moles
Sulfur = excess
FeS = 230 g
Molecular weight FeS = 88 g
Process
Reaction
Fe + S ⇒ FeS
1.- Convert the number of moles of Iron to grams
56 g of Iron -------------- 1 mol
x -------------- 3.8 moles
x = (3.8 x 56) / 1
x = 212.8 g of Iron
2.- Calculate the theoretical production of FeS
56 g of Iron ---------------- 88 g of FeS
212.8 g of iron ----------- x
x = (212.8 x 88) / 56
x = 334.4 g of FeS
3.- Calculate the percent yield
% yield = [tex]\frac{230}{334.4} x 100[/tex]
% yield = 68.8
Answer:
The percent yield of this reaction is 68.9 %
Explanation:
Step 1: Data given
Number of moles of iron = 3.8 moles
Iron reacts with excess of sulfur
230 grams of iron(II)sulfide is produced.
Step 2: The balanced equation
Fe + S → FeS
Step 3: Calculate moles FeS
For 1 mol Fe we need 1 mol S to produce 1 mol FeS
For 3.8 moles Fe we'll have 3.8 moles FeS
Step 4: Calculate mass of FeS
Mass FeS = moles FeS * molar mass FeS
Mass FeS = 3.80 moles * 87.91 g/mol
Mass FeS = 334 grams
Step 5: Calculate % yield
% yield = (actual yield / theoretical yield)*100%
% yield = (230/334) *100%
% yield = 68.9%
The percent yield of this reaction is 68.9 %
