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Question
Solve the following:
What volume of oxygen is required to burn completely 90 dm3 of butane under similar conditions of temperature and pressure?
\[\ce{2C4H10 + 13O2 -> 8CO2 + 10H2O}\]
Solution
\[\ce{\underset{2 Vols.}{2C4H10_{(g)}} + \underset{13 Vols.}{13O2}-> \underset{8 Vols.}{8CO2} + 10H2O}\]
2 vols. of butane requires O2 = 13 vols.
90 dm3 of butane will require O2 = `13/2 xx 90`
= 585 dm3
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