Calculate the mass (in g) of the excess reagent remaining after the reaction is complete.

Given 7.73 g of C5H10 and 6.47 g of O2, answer the following questions:
2 C5H10 + 15 O2 → 10 CO2 + 10 H2O

Calculate the mass (in g) of the excess reagent remaining after the reaction is complete.
I'm well on my way through this question but I can't figure out the correct mole ratio!

Calculate the theoretical yield of CO2 (in g) from the reaction.

Determine the percent yield of CO2 if 3.81 g of CO2 is actually produced from the reaction.

What is the limiting reagent?

Can someone please walk me through these question??

2 C5H10 + 15 O2 → 10 CO2 + 10 H2O
moles C5H10: 7.73 g/ 70 g/mole= 0.110
moles O2: 6.47 g/32 g /mole= 0.202
from bal. rxn., 15 moles of O2 required per 2 moles C5H10 (7.5:1)
for 0.110 moles C5H10, 0.825 moles O2 required(0.11 x 7.5), since there are only 0.202 moles O2, O2 is limiting reactant.
Excess reactant: 0.202 moles O2 /7.5=0.027 moles C5H10 reacted
0.110 - 0.027=0.083 moles C5H10 in excess
0.083 moles x 70 g/moles C5H10= 5.81 g C5H10 in excess
from bal. rxn., 10 moles of CO2 are formed from 15 moles of O2
theo. amt. of CO2: 0.202 moles O2 x 10/15 x 44 g/mole CO2= 5.93 g CO2
% Yield= actual/theo x100
% yield= 3.81/5.93 x 100
% yield= 64.2