# 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?

Mar 3, 2016

$19.2 m o l {O}_{2}$ or $615 g {O}_{2}$

#### Explanation:

The balanced equation of the combustion of ethane is:

$2 {C}_{2} {H}_{6} + 7 {O}_{2} \to 4 C {O}_{2} + 6 {H}_{2} O$

Then number of mole of oxygen that will be needed to completely react with $5.49 m o l$ 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