Liquid octane ch3ch26ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o. suppose 94. g of octane is mixed with 100. g of oxygen. calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. round your answer to 3 significant digits.

First, we write the balanced reaction equation:

C₈H₁₈ + 13.5O₂ → 8CO₂ + 9H₂O

From this, it is visible that the molar ratio between octane and oxygen should be 1 : 13.5

We check the actual amounts present by first converting them to moles:

Octane:

Mass = 94 g

Molar mass = 114 g/mol

Moles = 94/114 = 0.82

Oxygen:

Mass = 100 g

Molar mass = 32

Moles = 100/32 = 3.12

The molar ratio actually present is 1 : 3.81

This means that oxygen is the limiting reactant and we will base our calculations off of it.

3.12 moles of oxygen will produce:

3.12 * 8/13.5 = 1.85 moles of CO₂

The mass of CO₂ is calculated using:

Moles * molar mass

Mass = 1.85 * 44

The mass of CO₂ produced is 81.4 grams