Question

Calculate the volume of O2 required for complete combustion of 12.5 dm3 ethane gas at NTP ​

Answer 1

Answer:

first of all volume into litre.

1 dm3 = 1 litre

Answer 2

Answer:

43.75 dm³

Explanation:

Rate of combustion = 12.5dm³ (Given)

The combustion of ethane gas is described as -

C2H6 + (7/2)O2 →→ 2CO2 + 3H2O

where one mole of ethane gas is combused by 7/2 moles of oxygen gas.

Mole of oxygen at NTP gas -

1 × mole of the oxygen gas = 7/2 × mole of the ethane gas

1 × volume of oxygen gas/22.4 = 7/2 × volume of ethane gas/22.4

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

volume of oxygen gas = 7/2 × 12.5

volume of oxygen gas = 7/2 × 12.5

= 43.75 dm³

Thus, the volume of O2 required is 43.75 dm³.