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

2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

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Answer

[By Lussac’s law]

2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

2 vol    :  13 vol ⟶ 8 vol  : 10 vol

To calculate the volume of oxygen

C4H10    :     O2

2              :     13

90            :     x

⇒ x = 13\over 2 × 90 = 585 dm3

Hence, the volume of oxygen is required is 585 dm3

 

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