If 9.21 L of CH4 undergoes complete combustion at 73.04 mm Hg and 37.15ºC, how many grams of H2O is formed?CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

To solve this problem, we need to follow these steps:

1. Convert the volume of CH4 to moles using the ideal gas law. 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 from Celsius to Kelvin by adding 273.15. So, 37.15ºC = 310.3 K.

   Then, convert the pressure from mm Hg to atm because the R constant we're going to use requires the pressure to be in atm. 1 atm = 760 mm Hg, so 73.04 mm Hg = 73.04/760 = 0.0961 atm.

   Now, we can solve for n (number of moles of CH4): n = PV/RT = (0.0961 atm * 9.21 L) / (0.0821 Latm/molK * 310.3 K) = 0.359 moles of CH4.

2. Use the stoichiometry of the reaction to find the number of moles of H2O produced. According to the balanced chemical equation, 1 mole of CH4 produces 2 moles of H2O. So, 0.359 moles of CH4 will produce 0.359 * 2 = 0.718 moles of H2O.

3. Convert the moles of H2O to grams. The molar mass of H2O is approximately 18.015 g/mol, so 0.718 moles of H2O = 0.718 * 18.015 g/mol = 12.94 g of H2O.

So, if 9.21 L of CH4 undergoes complete combustion at 73.04 mm Hg and 37.15ºC, approximately 12.94 grams of H2O is formed.