Propane (C3H8) reacts with Oxygen gas to produce Carbon dioxide and water. What mass of CO2 will be produced if 8.5 liters of O2 reacts with 2.3 L of C3H8 @STP?

Question

Propane (C3H8) reacts with Oxygen gas to produce Carbon dioxide and water. What mass of CO2 will be produced if 8.5 liters of O2 reacts with 2.3 L of C3H8 @STP?

Answer 1

Hi Cindy,

First thing is we must write the correctly balanced equation for the reaction taking place:

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

Next, we use the stoichiometry of the balanced equation to find which reactant is limiting. One easy way to do this is to divide the mols or liters of each reactant by the corresponding coefficient in the balanced equation and whichever value is less that is the limiting reactant....

For C₃H₈: 2.3 L (÷1->2.3)

For O₂: 8.5 L (÷5->1.7)

Since 1.7 is less than 2.3, O₂ is the limiting reactant and will determine the amount of CO₂ that can form.

8.5 L O₂ x 3 L CO₂ / 5 L O₂ = 5.1 L CO₂ formed

Since the system is at STP, we know that 1 mol of gas = 22.4 L, so we can now convert 5.1 L CO₂ to mols CO₂ and then to grams CO₂ ...

5.1 L CO₂ x 1 mol / 22.4 L = 0.228 mols CO₂

0.228 mols CO₂ x 44 g CO₂ / mol = 10.0 g CO₂ = 10. g CO₂ (2 sig. figs.)