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?

1 Answer
Mar 3, 2016

Answer:

#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#