What volume of water vapor gas would be produced from the combustion of 510.16 grams of propane (C3H8) with 2,831.870 grams of oxygen gas, under a pressure of 0.96 atm and a temperature of 350. degrees C? Given: C3H8(g) + 5 O2lg) ---> 3 CO2(g) + 4 H20(g) (OR C3 H8 ("g") + 5 O2 ("g") right arrow 3 C O2 ("g") + 4 H2O ("g") 0.08206Latm mol K R Do not type units with your answer Your Answer:

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What volume of water vapor gas would be produced from the combustion of 510.16
grams of propane (C3H8) with 2,831.870 grams of oxygen gas, under a pressure of
0.96 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:
Transcribed Image Text:What volume of water vapor gas would be produced from the combustion of 510.16 grams of propane (C3H8) with 2,831.870 grams of oxygen gas, under a pressure of 0.96 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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