Question

calculate the volume of O2 required for complete combustion of 12.5 dm3 Ethane gas at ntp ​

Answer 1

answer : volume of oxygen gas = 43.75 dm³

explanation : combustion of ethane gas is given as....

C2H6 + (7/2)O2 ⇔2CO2 + 3H2O + (heat)

here it is clearly shown that for one mole of ethane gas is completely combused by (7/2) moles of oxygen gas.

or, (7/2) × mole of ethane gas = 1 × mole of oxygen gas

at NTP,

(7/2) × given volume of ethane gas/22.4L = 1 × volume of oxygen gas/22.4L

⇒(7/2) × 12.5 dm³ = volume of oxygen gas

⇒volume of oxygen gas = 7/2 × 12.5

volume of oxygen gas = 7/2 × 12.5= 43.75 dm³