100 cm³ of propane is mixed with 600 cm³ of oxygen and the mixture is ignited - Scottish Highers Chemistry - Question 18 - 2019

According to the balanced equation, 1 mole of propane reacts with 5 moles of oxygen. Thus, for 100 cm³ of propane, the required amount of oxygen is:

$100 ext{ cm}^3 imes 5 = 500 ext{ cm}^3$

Since we have 600 cm³ of oxygen, propane is the limiting reagent.

After the reaction, propane is consumed entirely, producing:

• Carbon dioxide (CO₂): 3 moles for every 1 mole of propane = 300 cm³ (as 1 mole can be approximated to 100 cm³ in volume)
• Water (H₂O): The water produced will be in vapor form, but it is excluded from the total gas volume since it condenses after cooling.

Total volume of gas after the reaction:

$ext{Total gas} = ext{Initial total} - ext{Used oxygen}$

Initial total gas = 100 cm³ (C₃H₈) + 600 cm³ (O₂) = 700 cm³

Used oxygen = 500 cm³

Total gas after reaction = 700 cm³ - 500 cm³ = 200 cm³ of remaining oxygen + 300 cm³ of carbon dioxide = 400 cm³.

Therefore, at the end of the reaction, the total volume of gas would be 400 cm³ (Option B).