Stoichiometry and chemical reactions

Question

How many grams of carbon dioxide, CO2 will be liberated when 35.0 grams of propane (C3H8) is burned with excess amount of oxygen, O2? (Molar mass of C=12.01 g/mol, H=1.01 g/mol, O=16.00 g/mol)

C3H8 + 5O2 = 3CO2 + 4H2O

Answer 1

First, write the correctly balanced equation for the combustion of propane:

C₃H₈ + 5O₂ ==> 3CO₂ + 4H₂O ... balanced equation

Since there is excess oxygen, we need not worry about limiting reactants, and the amount of CO₂ will be directly related to the amount (moles) of C3H8

Using molar mass, we first convert grams to moles, and then moles back to grams. We use the mole ratio of the balanced equation and dimensional analysis to solve this problem.

molar mass C₃H₈ = 44.1 g/mol

molar mass CO₂ = 44.0 g/mol

35.0 g C₃H₈ x 1 mol C₃H₈ / 44.1 g x 3 mol CO₂ / mol C₃H₈ x 44 g CO₂/mol = 105 g CO₂ (3 sig. figs.)