If 1.3 L of C3H8 combusts according to the equation below, how much CO2 will be produced?

C3H8 (g) + 5O2 (g) - -> 3CO2 (g) + 4H2O (g)

0.162 moles of CO₂ are produced by this reaction

Explanation:

The reaction is:

C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

As we have the volume of propane, we need to know the mass that has reacted, so we apply density's concept.

Density = Mass / Volume → Density. Volume = Mass

0.00183 g/mL. 1300 mL = Mass → 2.379 g

We determine the moles → 2.379 g. 1mol / 44 g = 0.054 moles

Ratio is 1:3. 1 mol of propane can produce 3 moles of dioxide

Then, 0.054 moles may produce (0.054.3) / 1 = 0.162 moles