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Question

 sample of 8.49 g of liquid 1‑propanol, C3H8O, is combusted with 55.1 g of oxygen gas. Carbon dioxide and water are the products.

Write the balanced chemical equation for the reaction

What is the limiting reactant?

How many grams of CO2 are released? 

How many grams of the excess reactant remain after the reaction is complete?

Answer 1

C₃H₈O + 5O₂ ==> 3CO₂ + 4H₂O ... balanced equation

To find the limiting reactant, one easy way is to simply divided the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

For C₃H₈O: 8.49 g x 1 mol / 60 g = 0.1415 moles (÷1->0.14)

For O₂: 55.1 g x 1 mol / 16 g = 3.44 moles (÷5->0.69)

Since 0.14 is < 0.69, this identifies C₃H₈O as the limiting reactant. We will now use moles of C₃H₈O to determine grams of CO₂ released:

0.1415 moles C₃H₈O x 3 moles CO₂ / 1 mol C₃H₈O x 44 g CO₂ / mole CO2 = 18.7 g CO₂