What volume of water vapor would be produced from the combustion of 595.39 grams of propane (C3H8) with 1,105.03 grams of oxygen gas, under a pressure of 1.033 atm and a temperature of 350. degrees C? Given: C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H20(g) ---> (OR C3 Hg ("g") + 5 O2 ("g") right arrow 3 C O2 ("g") + 4 H20 ("g") R = 0.08206Latm mol K Do not type units with your answer Your Answer:

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What volume of water vapor would be produced from the combustion of 595.39
grams of propane (C3H8) with 1,105.03 grams of oxygen gas, under a pressure of
1.033 atm and a temperature of 350. degrees C? Given:
C3H8(g) + 5 O2(g)
---> 3 CO2(g) + 4 H20(g)
(OR C3 H8 ("g") + 5 O2 ("g") right arrow 3 C O2 ("g") + 4 H20 ("g")
L atm
R = 0.08206
mol K
Do not type units with your answer
Your Answer:
Transcribed Image Text:What volume of water vapor would be produced from the combustion of 595.39 grams of propane (C3H8) with 1,105.03 grams of oxygen gas, under a pressure of 1.033 atm and a temperature of 350. degrees C? Given: C3H8(g) + 5 O2(g) ---> 3 CO2(g) + 4 H20(g) (OR C3 H8 ("g") + 5 O2 ("g") right arrow 3 C O2 ("g") + 4 H20 ("g") L atm R = 0.08206 mol K Do not type units with your answer Your Answer:
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