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

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the product of the pressure and volume is also constant