Question

Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to the following reaction. Molar Mass Fe2O3 = 159.7 g/mol

Answer 1

Answer:

83%

Explanation:

We first get the chemical reaction :

2 Al(s) + Fe2O3(s) ------> 2Fe(s) + Al2O3(s) 

From the reaction we can see that one mole of the oxide yielded 2 moles of iron.

Firstly, we need to calculate the theoretical yield of the iron. This is done as follows. The number of moles of the oxide equals the mass of the oxide divided by the molar mass of the oxide = 50g ÷ 159.7 = 0.313moles

From the first relation, one mole oxide yielded 2 moles iron, hence 0.313 mole oxide will yield 2 × 0.313 mole iron = 0.616 moles

The mass of iron thus generated = 0.616 × 56 = 34.496g

% yield = Actual yield/theoretical yield × 100%

%yield = 28.65/34.396 × 100% = 83%

Answer 2

Answer:

The percent yield of the reaction is 82%

Explanation:

First step: make the chemist equation.

2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)

As the statement says that aluminun is in excess, the limiting reactant is the Fe2O3

Second step: Find out the moles in the reactant.

Molar weight Fe2O3: 159.7 g/m

Mass / Molar weight = moles

50 g /159.7 g/m = 0.313 moles

Third step: Analyse the reaction. 1 mol of Fe2O3 makes 2 moles of Fe.

1 mol Fe2O3 ____ 2Fe

0.313 mol Fe2O3 ____ 0.626 moles

Molar weight Fe = 55.85 g/m

Moles . molar weight = mass

55.85g/m . 0.626m = 34.9 grams

This will be the 100% yield of the reaction but we only made 28.65 g

34.9 g ____ 100%

28.65 g ____ 82.09 %