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Chemistry

What volume of oxygen is required to burn completely 90 dm3 of butane under similar conditions of temperature and pressure?

2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

Stoichiometry

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Answer

[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}

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}

∴

132×90=585 dm3\dfrac{13}{2} \times 90 = 585 \text{ dm}^3

Hence, volume of oxygen is required is 585 dm3

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