Discover the Exciting World of Chemistry: Grams of CO2 Produced in Combustion

How can we calculate the grams of CO2 produced in the complete combustion of a 406-g bottled gas?

Given that the bottled gas consists of 72.7% propane, what steps do we need to follow to determine the amount of CO2 produced?

Calculating Grams of CO2 in Complete Combustion

To find the grams of CO2 produced in the complete combustion of a 406-g bottled gas that is 72.7% propane, follow these steps:

First, calculate the moles of propane, then apply stoichiometry to determine the moles of CO2 produced, and finally convert to grams.

Exploring the Calculation Process

When dealing with the complete combustion of a 406-g bottled gas containing 72.7% propane (C3H8), the process begins by calculating the moles of propane present. Since the molecular weight of propane is around 44.09 g/mol, we can compute the moles as follows:

406g * 72.7% / 44.09 g/mol = 6.67 mol of propane

According to the combustion reaction of propane with oxygen, it yields water and carbon dioxide: C3H8 + 5O2 -> 3CO2 + 4H2O. This equation signifies that for every mole of propane, three moles of CO2 are produced. Hence, for 6.67 mol of propane, the calculation results in:

6.67 mol * 3 = 20.01 mol of CO2

To convert the moles of CO2 into grams, we utilize the molecular weight of CO2, approximately 44.01 g/mol, leading to:

20.01 mol * 44.01 g/mol = 880.5g of CO2 produced