Question

I'm having trouble solving these kind of problems. I can't quite figure out which is the limiting reactant when the problem only gives grams and not mols. How do I solve this step by step?

Ethene, C₂H₄, burns in air according to the equation below. What is the theoretical yield (in grams) of CC₂H₄,when 3.24 g of C₂H₄ are reacted with 4.83 g of O₂

C₂H₄ + 3O₂ - - -> 2CO₂ + 2H₂O

A.9.72 g

B.

0.151 g

C.

4.43 g

D.

3.24 g

Answer 1

Answer: C. 4.43 g

C₂H₄ + 3O₂ - - -> 2CO₂ + 2H₂O

1 mol C2H4 reacts with 3 mol O2 and gives 2 mol CO2 and 2 mol H2O

moles C2H4 = mass of C2H4 / molar mass of C2H4 = 3.24 g / 28.05 g/mol = 0.11551 mol

moles O2 = mass of O2/ molar mass of O2 = 4.83 g / 31.9988 g/mol = 0.1509 mol

0.11551 mol C2H4 requires = 3 x 0.11551 mol O2 = 0.34653 mol O2, but we do not have such amount, so O2 is limiting reactant

0.1509 mol O2 reacts with only 0.1509 / 3 = 0.0503 mol C2H4 and produces 0.0503 x 2 = 0.1006 mol CO2 and 0.1006 mol H2O

convert mol CO2 to mass CO2 by multiplying mol CO2 with molar mass of CO2 -----> this is theoretical yield of CO2

Theoretical yield of CO2 = 0.1006 mol x 44.01 g/mol = 4.427406 g ~ 4.43 g

Theoretical yield of CO2 = 4.43 g

Hope this helped you!

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