What volume of oxygen is required to burn completely 90 dm3 of butane under similar conditions of temperature and pressure?
2C4H10 + 13O2 ⟶ 8CO2 + 10H2O
[By Lussac's law]
2C4H10+13O2⟶8CO2+10H2O2 vol.:13 vol.⟶8 vol.:10 vol.\begin{matrix} 2\text{C}_4\text{H}_{10} & + & 13\text{O}_2 & \longrightarrow & 8\text{CO}_2 & + & 10\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 13 \text{ vol.} & \longrightarrow & 8\text{ vol.} & : & 10\text{ vol.} \end{matrix}2C4H102 vol.+:13O213 vol.⟶⟶8CO28 vol.+:10H2O10 vol.
To calculate the volume of oxygen
C4H10:O22:1390:x\begin{matrix}\text{C}_4\text{H}_{10} & : & \text{O}_2 \\ 2 & : & 13 \\ 90 & : & x \end{matrix}C4H10290:::O213x
∴
132×90=585 dm3\dfrac{13}{2} \times 90 = 585 \text{ dm}^3213×90=585 dm3
Hence, vol of oxygen = 585 dm3