A sample of 7.64 g of liquid 1‑propanol, C3H8O, is combusted

Question

A sample of 7.64 g of liquid 1‑propanol, C3H8O, is combusted with 61.3 g of oxygen gas. Carbon dioxide and water are the products.

Write the balanced chemical equation for the reaction. Physical states are optional.

a.) what is the limiting reactant

b.) How many grams of CO2 are released in the reaction?

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

Answer 1

2C₃H₈O + 9O₂ ==> 6CO₂ + 8H₂O ... balanced equation

(a). To find the limiting reactant we can divide the mols of each reactant by the corresponding coefficient in the balanced equation. Whichever answer is less represents the limiting reactant.

C₃H₈O: 7.64 g x 1 mol / 60.1 g = 0.127 mols (÷2->0.06)

O₂: 61.3 g x 1 mol / 32 g = 1.92 mols (÷9->0.21)

Since 0.06 is less than 0.21, C₃H₈O is the limiting reactant

(b). Use the mols of the limiting reactant (C₃H₈O) to find grams CO₂ formed:

0.127 mol C₃H₈O x 6 mol CO₂ / 2 mol C₃H₈O x 44 g CO2 / mol = 16.8 g CO₂ formed

(c). Excess reactant is O₂, so find out how much was used and then subtract that from what we started with to find out how much is left over.

0.127 mol C₃H₈O x 9 mol O₂ / 2 mol C₃H₈O x 32 g O₂ / mol = 18.3 g O₂ used up

61.3 g O₂ - 18.3 g O₂ = 43.0 g O₂ left over