What is the theoretical yield of Cl2? - Show Work Please

Question

Question: If 18.3 grams of C2H5 were reacted with 37.3 grams of O2, how many grams of Cl2 would be produced?

 

Equation: 4C2H5Cl + 13O2 = 8CO2 + 10H2O + 2Cl2 (already balanced)

Answer 1

** I think you were missing a chloride in C₂H₅ in question.

 

1. Calculate the molecular weight of each reagent/product. Use it to convert all given mass quantities to mole.

 

* C₂H₅Cl

2(12 g/mol) + 5(1 g/mol) + 1(35.5 g/mol)  = 64.5 g/mol C₂H₅Cl

 

18.3 grams of C₂H₅Cl

(18.3 g) / (64.5 g/mol ) = 0.284 mol C₂H₅Cl

 

O₂

2(16 g/mol) = 32 g/mol O₂

 

37.3 grams of O₂

(37.3 g)/(32 g/mol) = 1.17 mol O₂

 

Cl₂

2(35.5 g/mol) = 71 g/mol Cl₂

 

2. Determine the limiting reagent using the calculated moles and stoichiometry of the reaction.

 

4 C₂H₅Cl will make 2 Cl₂

0.284 mol C₂H₅Cl * 2 mol Cl₂ / 4 mol C₂H₅Cl = 0.142 mol Cl₂

 

13 O₂ will make 2 Cl₂

1.17 mol O₂ * 2 mol Cl₂ / 13 mol O₂ = 0.18 mol Cl₂

 

The limiting reagent is C₂H₅Cl because we aren't able to make as much product.

0.18 g/mol > 0.142 g/mol

 

0.142 mol Cl₂ is maximum that can be produced given the starting quantities of reagents.

 

3. calculate grams product from the mole product produced using the limiting reagent.

 

0.142 mol Cl₂ * 71 g/mol Cl₂ = 10.01 g Cl₂