Given that the molar mass of h2o is 18.02 g/mol, how many liters of propane are required at stp to produce 75 g of h2o from this reaction?
23 l23.3 l93 l93.2 l

First, we require the combustion equation of propane:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

We see that each mole of propane produces 4 moles of water.
Now, we calculate the moles of water required: 75 / 18.02
Moles required = 4.16
Moles of propane required for this are: 4.16 / 4 = 1.04 moles

Each mole of gas occupies 22.14 liters, so the volume of propane required are 1.04 * 22.14 = 23.03 L
23 L of propane are required