Suppose 22. g of octane is mixed with 98.0 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction.

Answer 1

First, you must write a correctly balanced equation for the reaction taking place:

octane = C₈H₁₈

Oxygen = O₂

2C₈H₁₈ + 25O₂ ===> 16CO₂ + 18H₂O ... balanced equation for combustion of octane

To find the maximum mass of CO2, we need to look at the balanced equation and note that 2 MOLES of octane reacts with 25 MOLES of oxygen to produce 16 MOLES of CO₂.

In this problem we are given the GRAMS of octane and oxygen. We must convert them to moles, and then find which one (octane or oxygen) is limiting.

moles octane = 22 g x 1 mole/114 g = 0.19 moles

moles oxygen = 98.0 g x 1 mole/32 g = 3.1 moles

Since the ratio of O₂ to octane is 25 : 2 we can divide 0.19 by 2 to get 0.095 and divide 3.1 by 25 to get 0.12, so OCTANE is limiting and will dictate how much CO₂ can be produced

mass of CO₂ produced: 0.19 moles C₈H₁₆ x 16 moles CO₂ / 2 moles C₈H₁₈ x 44 g CO₂/mol = 67 g CO₂ (to 2 significant figures)