If 5.00 mol of ethane, C2H6, undergoes combustion, according to the unbalanced equation given, how many moles of oxygen are required: C2H6 + O2 ---> CO2 + H2O?
Your question is a little ambiguous; if you use the unbalanced equation, the way you have it written, one mole of ethane will use one mole of oxygen. So if you have five moles of ethane, since the ratio is 1:1 in the unbalanced equation, you would require five moles of oxygen to react with the five moles of ethane.
However, if you balance the equation, the numbers work out like this:
2C2H6 + 7O2 ---> 4CO2 + 6 H2O
What this means is for every mole of ethane, you have to have three and one-half moles of oxygen to react with it. So if you have five moles of ethane, by this ratio we will need 3.5 times that number of oxygen to react with it, or seventeen and one-half moles of oxygen to react with 5 moles of ethane. Or....if you want to eliminate the halves, five moles of ethane times the 2 of the balanced equation would give you 10 moles of ethane. Five times the 7 of the oxygen in the balanced equation would give you 35 moles of oxygen. Either way, it still gives you the same ratio that is demonstrated in the balanced equation.