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Chemistry

What volume of oxygen is required to burn completely a mixture of 22.4 dm3 of methane and 11.2 dm3 of hydrogen into carbon dioxide and steam?

CH4 + 2O2 ⟶ CO2 + 2H2O

2H2 + O2 ⟶ 2H2O

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Answer

CH4+2O2CO2+2H2O1 vol.:2 vol.1 vol.\begin{matrix} \text{CH}4 & + & 2\text{O}2 & \longrightarrow & \text{CO}2 & + & 2\text{H}2\text{O} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1\text{ vol.} \\ \end{matrix}

From equation:

22.4 dm3 of methane requires oxygen = 2 x 22.4 dm3 of O2 = 44.8 dm3

2H2+O22H2O2 vol.:1 vol.2 vol.\begin{matrix} 2\text{H}2 & + & \text{O}2 & \longrightarrow & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \\ \end{matrix}

From equation,

[2 x 22.4] dm3 hydrogen requires oxygen = 22.4 dm3

∴ 11.2 dm3 hydrogen will require oxygen = 22.42×22.4\dfrac{22.4}{2 \times 22.4} x 11.2 =

= 5.6 dm3

Total volume of oxygen required = 44.8 + 5.6 = 50.4 dm3

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