[By Lussac’s law]

2C_{2}H_{2} + 5O_{2 } → 4CO2 + 2H_{2}O

2 vol : 5 vol → 4 vol : 2 vol

**To calculate the volume of air required,**

C_{2}H_{2} : O_{2}

2 : 5

50 : x

Therefore, volume of oxygen (x),

⇒ x = 50 × 5\over 2 = 125 cm^{3}

When oxygen is 20% then air is 100%

Therefore when, oxygen is 125 cm^{3} then air is

⇒ x = 125 × 100\over 20 = 625 cm^{3}

Therefore, **volume of air required is 625 cm ^{3}**.