What is the percent yield of a reaction in which 3.8 moles of iron metal react with excess sulfur to produce 230 grams of iron(ii)sulfide? fe + s --> fes

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 =

% yield = 68.8

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 %

just did the assignment and the answer was #1 +2 n

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