When 45.8 g C3H8 reacts with 215 g O2, how many grams of CO2 can be made?

Question

A. 137

B. 59.1

C. 45.7

D. 177

Answer 1

Limiting reactant problem:

Write the correctly balanced equation:

C₃H₈ + 5O₂ ==> 3CO₂ + 4H₂O ... balanced equation

moles C₃H₈ present = 45.8 g x 1 mol C₃H₈/44.1 g = 1.039 moles

moles O₂ present = 215 g O₂ x 1 mole/32 g = 6.719 moles

Since it takes 5 moles of O₂ for each 1 mole C₃H₈ (see balanced equation), C₃H₅ is limiting. You can easily see this by dividing moles of each by their respective coefficient. Thus...

for C₃H₈ you have 1.039 / 1 = 1.039

for O₂ you have 6.719 / 5 = 1.34 which is greater than 1.039 so C₃H₈ is limiting

Find mass of CO₂ that can be made:

Using the limiting reactant as the determining factor, we have...

1.039 mol C₃H₈ x 3 mol CO₂/mol C₃H₈ x 44 g CO₂/mol CO₂ = 137 g CO₂ formed (theoretically)