A gas grill burns Propane (C3H8) in the presence of more than sufficient Oxygen (O2). This reaction produces water vapor and Carbon Dioxide. The temperature and pressure conditions are such that 1 mole of each gas occupies 1 liter of volume.
If 15 liters of Propane are completely consumed, how many liters of water vapor will be produced?
A. 7.5 L H2O (g)
B. 15 L H2O (g)
C. 30 L H2O (g)
D. 45 L H2O (g)
E. 60 L H2O (g)
1
Expert's answer
2018-01-22T05:52:29-0500
1L 4L С3Н8 + 5О2 = 3СО2 + 4Н2О 15L x L 1L/15L = 4L/x L x=15*4 / 1=60 L E. 60 L H2O (g)
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