If 116g of butane burns in presence of 320g of oxygen, how much carbon dioxide will be produced? What is limiting reagent?

Question

So This is how I worked it out:
116gC4H10*1/58.12(molar mass of C4H10)*8 moles CO2/2 moles C4H10*44/1(molar mass of CO2)=351.27g of CO2

Answer 1

Is this answer correct? I looked up in the computer and it said the answer is supposed to be 270.84g of CO2 but I thought I did this problem right.

Answer 2

Oxygen is the limiting reagent in the problem. 320g molecular oxygen = 10 mole oxy. solve (13/10)moles oxy = (8/x)moles CO2. x = 6.15 moles CO2 x 44g/mole CO2 = 270.8 gms CO2

Answer 3

Thank You!!