Question

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?

Answer 1

`19.2molO_2` or `615gO_2`

Explanation:

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`