If 5.49 mol of ethane ($C_2H_6$) undergoes combustion according to the unbalanced equation $C_2H_6+O_2->CO_2+H_2O$, how much oxygen is required?

Md Ariful Islam

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Md Ariful Islam

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The balanced equation of the combustion of ethane is:

$2C_2H_6+7O_2->4CO_2+6H_2O$

Then number of mole of oxygen that will be needed to completely react with $5.49mol$ of ethane could be calculated as follows:

$?molO_2=5.49 cancel(mol C_2H_6)xx(7molO_2)/(2cancel(molC_2H_6))=19.2molO_2$

The mass of oxygen needed would be:

$?gO_2=19.2 cancel(molO_2)xx(32.0gO_2)/(1cancel(molO_2))=615gO_2$

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