Ideal gas law problem

Question

1. How many grams of water would be produced if 20.0 liters of oxygen were burned at a temperature of -10.0 C and a pressure of 1.3 atm. Use the following equation

                            2 C8 H18 +25 O2 --> 16CO2+18H2O

 

 

 

 

Answer 1

Hello, Emily,

This problem requires two independent steps. We'll use the idel gas law to determine the number of moles of O₂, which we can then use in the second step of cacluating the grams of water produced. We'll assume we have an excess of C₈H₁₈.

Ideal Gas Law

We need to find the moles of O₂ contained in the 20.0 L of oxygen. We need the ideal gas law:

PV = nRT

where n is the moles and R the gas constant. We are given gas conditions of 1.3atm and -10.0 C. The gas laws reqire Kelvin for temperature, so add 273.15 to the -10C to give us 263.15K. The next important decision is to chose a value of R, the gas constant, that has as many of the same units as we were given (atm, K, and liters). This simply reduces the number of conversions we need so that the units cancel and leave just moles. I used R= 0.0821 L*atm*K^-1mol^-1.

Rearrange the equation to isolate the unknown, and then enter the data:

n = (1.3atm)(20.0L)/((0.0821 L*atm*K^-1mol^-1)*(263.2K))

Cross out the units that cancel, and check to be sure we're left with only moles. (It's fun with pencil, but a pain with a mouse). I think we're OK - I get 1.20 moles O₂.

Balanced Equation

We have 1.20 moles of O_2, an excess of C₈H₁₈, and a balanced equation:

2 C₈H₁₈ + 25 O₂ = 16 CO₂ + 18H₂O

This tells us we'll get 18 moles of water for every 25 moles of oxygen, a molar ratio of 18/25.

This means we'd expect (1.20 moles O₂)*(18 moles H₂O/25 moles O₂) or 0.867 moles H₂O. Convert to grams by multiplying by water's molar mass: (0.867 moles)*(18 g/mole) = 15.6 grams.

That was fun, and I think it is correct.

Bob