How many grams of water would be produced if 2.45 liters of oxygen were burned at a temperature of 95.743°C and a pressure of 7.54 atm?

To answer this question, we need to use the ideal gas law and stoichiometry.

1. First, we need to convert the volume of oxygen from liters to moles. 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. We can rearrange the equation to solve for n: n = PV/RT.

2. Convert the temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is K = °C + 273.15. So, 95.743°C = 95.743 + 273.15 = 368.893 K.

3. The ideal gas constant (R) is 0.0821 L·atm/(K·mol) when pressure is in atm and volume is in liters.

4. Now we can plug in the values into the equation: n = (7.54 atm * 2.45 L) / (0.0821 L·atm/(K·mol) * 368.893 K) = 0.61 moles of O2.

5. The balanced chemical equation for the combustion of hydrogen is 2H2 + O2 -> 2H2O. This means that one mole of oxygen produces two moles of water.

6. Therefore, 0.61 moles of O2 would produce 2 * 0.61 = 1.22 moles of H2O.

7. Finally, we convert moles of water to grams. The molar mass of water is approximately 18.015 g/mol. So, 1.22 moles * 18.015 g/mol = 21.98 g.

So, if 2.45 liters of oxygen were burned at a temperature of 95.743°C and a pressure of 7.54 atm, approximately 21.98 grams of water would be produced.