#### Question

Propane burns in air according to the following equation:

C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O

What volume of propane is consumed on using 1000 cm^{3} of air, considering only 20% of air contains oxygen?

#### Solution

Given:

C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O

Volume of air = 1000 cm^{3}

Percentage of oxygen in air = 20%

From the given information

C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O

1vol 5vols 3vols 4vols

According to Gay-Lussac’s law,

1 vol. of propane consumes 5 vol. of oxygen

Volume of oxygen = 1000 cm^{3} × 20% = 200 cm^{3}

Therefore,

Volume of propane burnt for every 200 cm^{3} of oxygen

`= 1/5 xx 200 = 40 cm^3`

40 cm^{3} of propane is burnt.

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#### APPEARS IN

Solution What Volume of Propane is Consumed on Using 1000 Cm3 of Air, Considering Only 20% of Air Contains Oxygen? Concept: Gay Lussac’s Law of Combining Volumes.