Identify the limiting reactant and determine the mass of CO2 that can be produced from the reaction of 25.0 g of C3H8 with 2.3 moles of O2 according to he following equation: C3H8 + 5O2 = 3CO2 + 4H2O

Question

As far as I understand it, I must divide the number of moles of each reaction by its stoichiometric coefficient to identify limiting reactants. But every time I do so I get a different answer. 

Answer 1

You are correct that you must divide the moles of each reactant by the corresponding coefficient. This is illustrated below.

First, write the correctly balanced equation:

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

Find limiting reactant:

For C₃H₈: 25 g C₃H₈ x 1 mol C₃H₈/ 44g = 0.568 mols C₃H₈ (÷1->0.568)

For O₂: 2.3 mols O₂ (÷5->0.46)

Since 0.46 is less than 0.568, this means O₂ is the limiting reactant

Now, to determine the mass of CO₂ that can be produced, we must use the moles of the limiting reactant.

mass CO₂ = 2.3 mols O₂ x 3 mols CO₂ / 5 mols O₂ x 44 g CO₂/mol = 60.7 g CO₂ produced