If 6.15 L of CH4 undergoes complete combustion at 943.97 mm Hg and 83.47ºC, how many grams of H2O is formed?CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

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

1. Convert the temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is K = C + 273.15. So, 83.47ºC = 83.47 + 273.15 = 356.62 K.

2. Convert the pressure from mm Hg to atm. The conversion factor is 1 atm = 760 mm Hg. So, 943.97 mm Hg = 943.97 / 760 = 1.2423 atm.

3. Use the ideal gas law (PV = nRT) to find the number of moles of CH4. In this case, P = 1.2423 atm, V = 6.15 L, R = 0.0821 L·atm/K·mol (the ideal gas constant), and T = 356.62 K. Solving for n (the number of moles), we get n = PV / RT = (1.2423 atm * 6.15 L) / (0.0821 L·atm/K·mol * 356.62 K) = 0.262 moles of CH4.

4. Use the balanced chemical equation to find the number of moles of H2O produced. According to the equation, 1 mole of CH4 produces 2 moles of H2O. So, 0.262 moles of CH4 will produce 0.262 * 2 = 0.524 moles of H2O.

5. Convert the number of moles of H2O to grams. The molar mass of H2O is approximately 18.015 g/mol. So, 0.524 moles of H2O = 0.524 * 18.015 g/mol = 9.44 g of H2O.

So, if 6.15 L of CH4 undergoes complete combustion at 943.97 mm Hg and 83.47ºC, approximately 9.44 grams of H2O is formed.