In the combustion of 250 g of propane with an excess of hydrogen, what volume of carbon dioxide is produced at 290 K and 1.5 atm. Choose from the following possible answers:
11904.5 L
270 L
90 L
404 L
2007-11-20 10:14:31 · 2 answers · asked by Goddess Allure 2 in Science & Mathematics ➔ Chemistry
Hydrogen? Or oxygen?
Anyway, convert g propane to mol propane, work out how many mol CO2 is produced from each mole propane (not a difficult question, since each carbon in propane gives you just one CO2), and then use
PV = nRT
That's all!
2007-11-20 10:21:54 · answer #1 · answered by Facts Matter 7 · 0⤊ 0⤋
You have to mean combustion with OXYGEN, right?
C3H8 + 5 O2 -------> 3 CO2 + 4H2O
Moles C3H8 = 250 g / 44g/ mole = 5.68 mole
Moles CO2 produced = 3 X 5.68 mole = 17.0 mole
PV = nRT
V= nRT / P = (17mole)(0.0821 l atm/mole K)(290K) / 1.5 atm = 270 liters
Answer: 270 liters
2007-11-20 10:25:10 · answer #2 · answered by Dennis M 6 · 0⤊ 0⤋