Question

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)

Answer 1

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)