The combustion of propane may be described by the chemical equation,
C3H8(g)+5O2(g)——>3CO2(g)+4H2O(g)
How many grams of O2(g) are needed to completely burn 59.9 g of C3H8(g)?
_____g O2
2 Answers

mole C3H8= 44 gm
mole O2= 32 gm
44 gm C3H8 reacts with 5× 32 gm O2
59.9 gm C3H8 reacts with X gm O2
O2=5×32×59.9 ÷44=217.818 gm

One mole of C3H8 weighs (3*12 + 8*1 = 44 grams.
That means 59.9 grams represents 59.9 grams / 44 grams/mol = 1.361 moles.
Now the balanced equation tells us that you need 5 moles of O2 to completely burn 1 mole of C3H8
So we need 5 times 1.361 moles of O2 which is 6.805 moles of O2.
Lastly how much does 6.805 moles of O2 weigh?
Now 1 mole of O2 weighs 16*2 = 32 grams per mole.
So 6.805 moles of O2 weighs 6.805 * 32 = 217.76 grams.