In the combustion reaction of 149g of propane(C3H8) with excess oxygen, what volume of carbon dioxide(CO2) is produced at STP?
(note: balance the reaction first)
C3H8 + O2 3CO2 + H2O
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The balanced reaction is:
C3H8 + 5O2 3CO2 + 4H2O
Number of moles of propane is 149g/44g/mole = 3.386 moles
Ratio of moles C3H8:CO2 is 1:3.
So we can use the proportion: 1/3=3.386/x, where x is a number of moles of CO2.
x = 3.386*3 = 10.158 moles of CO2
Volume of CO2 is 10.158 moles*22.4L = 227.54L
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