Answer: 11.2 moles of [tex]CO_2[/tex] are produced when 5.60 mol of ethane is burned in an excess of oxygen.
Explanation:
The combustion of ethane is represented using balanced chemical equation:
[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)[/tex]
As oxygen is preset in excess, ethane acts as the limiting reagent as it limits the formation of product.
According to stoichiometry :
2 moles of propane produces 4 moles of carbon dioxide
Thus 5.60 moles of propane will produce=[tex]\frac{4}{2}\times 5.60=11.2[/tex] moles of carbon dioxide
Thus 11.2 moles of [tex]CO_2[/tex] are produced when 5.60 mol of ethane is burned in an excess of oxygen.
