36.7 grams of FeS2 reacts with 50.9 grams of O2 by the reaction below.

4 FeS2 +11 O2 → 8 SO2 + 2 Fe2O3

The molar masses of the reaction components are:

FeS2: 119.98 g/mol

O2: 32.00 g/mol

Fe2O3: 159.69 g/mol

SO2: 64.06 g/mol

1) What is the limiting reactant?

2)What is the theoretical yield of the second product listed in the reaction above?

1) The limiting reactant is FeS₂.

 

2) The theoretical yield of Fe₂O₃ is 24.4 g.

Step-by-step explanation

Given:

Mass of FeS₂: 36.7 g

Mass of O₂: 50.9 g

 

Molar masses:

FeS₂: 119.98 g/mol

O2: 32.00 g/mol

Fe2O3: 159.69 g/mol

SO2: 64.06 g/mol

 

The balanced equation showing the reaction that takes place has been provided as:

4 FeS₂ + 11 O₂ ==>8 SO₂ + 2 Fe₂O₃

 

 

1) Limiting reactant

 

The limiting reactant is the reactant that is completely used up in a reaction. It is said to be limiting because a reaction stops as soon as all the limiting reactant has reacted. As a result, the limiting reactant determines how much product is formed.

 

To determine which reactant is the limiting reactant, calculate the amount of product(Fe₂O₃) that would be formed from each respective reactant. The reactant that forms the least amount of product is the limiting reactant.

 

Step 1: Convert the mass of each reactant into its respective moles

 

Moles = Mass/Molar Mass

 

Mol FeS₂ = Mass of FeS₂/Molar mass of FeS₂ 

= 36.7 g/119.98 g.mol^-1

= 0.3059 mol

 

 

Mol O₂ = Mass of O₂/Molar mass of O₂ 

= 50.9 g/32.00 g.mol^-1

= 1.5906 mol

 

Step 2: Calculate the amount of Fe₂O₃ that would be formed from each reactant.

 

a) Amount of Fe₂O₃ that would be formed from FeS₂

 

From the balanced equation, the mol ratio of FeS₂ to Fe₂O₃ is 4:2 which when simplified becomes 2:1. This means that for every 2 moles of FeS₂ consumed in the reaction, only 1 mole of Fe₂O₃ would be formed.

 

Thus:

Mol Fe₂O₃ = 1/2 x Mol FeS₂ = 1/2 x 0.3059 mol = 0.1530 mol

 

 

b) Amount of Fe₂O₃ that would be formed from O₂

 

From the balanced equation, the mol ratio of O₂ to Fe₂O₃ is 11:2. This means that for every 11 moles of O₂ consumed in the reaction, only 2 moles of Fe₂O₃ would be formed.

 

Thus:

Mol Fe₂O₃ = 2/11 x Mol O₂ = 2/11 x  1.5906 mol = 0.2892 mol

 

 

From the calculations above:

• FeS₂ would form 0.1530 moles of Fe₂O₃
• O₂ would form 0.2892  moles of Fe₂O₃

 

Therefore, FeS₂ is the limiting reactant because it leads to the formation of the least amount of product(Fe₂O₃).

 

 

2) Theoretical yield of the second product(Fe₂O₃).

Theoretical yield is the maximum possible amount of product that can be formed from the given reactants. In other words, it is the amount of product formed assuming all the limiting reactant is completely used up in a reaction.

From the calculations above, it has been determined that FeS₂ is the limiting reactant.

It has also been determined that FeS₂ would form 0.1530 mol of Fe₂O₃.

Convert the mass of Fe₂O₃ to mass as follows:

Mass = Moles x Molar mass

Mass of Fe₂O₃ = Moles of Fe₂O₃ x Molar mass of Fe₂O₃

= 0.1530 mol x 159.69 g/mol

= 24.4 g

Therefore, the theoretical yield of Fe₂O₃ is 24.4 g.