Question

Using the following equation for the combustion of octane, calculate the amount of grams of carbon dioxide formed from 100.0 g of octane. The molar mass of octane is 114.33 g /mole .The molar mass of carbon dioxide is 44.0095 g/mole.

2 C8H18 +25 O2 ---->16 CO2 +18 H20
/\H 11018 kJ

Answer 1

307.94g of CO₂

Step-by-step explanation:

Equation of reaction

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

molarmass of C₈H₁₈ = 114.33g/mol

2 moles of C₈H₁₈ = 228.66g/mol

Molar mass of CO₂ = 44.0095g/mol

16 moles of CO₂ = 704.152g/mol

From equation of reaction,

2 moles of C₈H₁₈ reacts with 16 moles of CO₂;

228.66g of C₈H₁₈ = 704.152g of CO₂

100g of C₈H₁₈ = y moles of CO₂

y = (100 * 704.152) / 228.66

y = 307.94g of CO₂

Therefore, 100g of C₈H₁₈would yield 307.94g of CO₂.

Answer 2

The balanced chemical reaction is written as:

2C8H18 +25O2 ---->16CO2 +18H20

We are given the amount of octane to be used in the reaction. We use this as the starting point of our calculations. We go as follows:

100.0 g C8H18 ( 1 mol / 114.33 g ) (16 mol CO2 / 2 mol C8H18 ) (44.0095 g / mol ) = 307.95 g CO2 is produced