2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g)How many grams of water would be produced if 1.04 liters of oxygen were burned at a temperature of -12.431°C and a pressure of 7.09 atm?

To solve this problem, we need to use the ideal gas law and stoichiometry.

Step 1: Use the ideal gas law to find the number of moles of oxygen. The ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.

First, convert the temperature to Kelvin. The formula to convert Celsius to Kelvin is K = °C + 273.15. So, -12.431°C = -12.431 + 273.15 = 260.719 K.

Next, we need to convert the pressure to atmospheres to match the units of the ideal gas constant. The pressure is already given in atmospheres, so no conversion is necessary.

Now we can plug the values into the ideal gas law:

(7.09 atm) * (1.04 L) = n * (0.0821 Latm/Kmol) * (260.719 K)

Solving for n, we get n = 0.035 moles of O2.

Step 2: Use stoichiometry to find the number of moles of water produced. According to the balanced chemical equation, 25 moles of O2 produce 18 moles of H2O. So, 0.035 moles of O2 would produce (0.035 moles O2) * (18 moles H2O / 25 moles O2) = 0.0252 moles of H2O.

Step 3: Convert moles of water to grams. The molar mass of water is approximately 18.015 g/mol. So, 0.0252 moles of H2O * 18.015 g/mol = 0.454 grams of H2O.

So, if 1.04 liters of oxygen were burned at a temperature of -12.431°C and a pressure of 7.09 atm, approximately 0.454 grams of water would be produced.