Sample 5.26 g of liquid 1‑propanol,C3H8O,is combusted with 31.8 g of oxygen gas.Carbon dioxide and water are the products. How many grams of the excess reactant remain after the reaction is complete?

Answer 1

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

moles propanol = 5.26 g x 1 mol/60.1 g = 0.0875 moles

moles O₂ = 31.8 g x 1 mol/31.9 g = 0.997 moles O₂

Propanol is limiting based on the mol ratio in balance equation of 2 : 9

To find mass of O₂ (excess reagent) left over, we will first find moles O₂ used up.

moles O₂ used = 0.0875 mol propanol x 9 mol O₂/2 mol propanol = 0.394 moles O₂ used

moles O₂ left over = 0.997 mol - 0.394 mol = 0.603 mol O₂ left

mass O₂ left = 0.603 mol O₂ x 32 g/mol = 19.3 g O₂ left over