Question

A sample of 5.72 g of liquid 1-propanol, C, H, O, is combustod with 43.4 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