[By Lussac’s law]

2C_{4}H_{10} + 13O_{2} ⟶ 8CO_{2} + 10H_{2}O

2 vol : 13 vol ⟶ 8 vol : 10 vol

To calculate the volume of oxygen

C_{4}H_{10} : O_{2}

2 : 13

90 : x

⇒ x = 13\over 2 × 90 = 585 dm^{3}

Hence, the **volume of oxygen is required is 585 dm ^{3}**