If 6.15 L of CH4 undergoes complete combustion at 943.97 mm Hg and 83.47ºC, how many grams of H2O is formed?

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

Step 1: Convert the given conditions to the appropriate units for the ideal gas law.

The ideal gas law is PV = nRT, where P is pressure in atmospheres, V is volume in liters, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/K·mol), and T is temperature in Kelvin.

Given: P = 943.97 mm Hg = 943.97 mm Hg * (1 atm / 760 mm Hg) = 1.242 atm (approx) V = 6.15 L T = 83.47ºC = 83.47 + 273.15 = 356.62 K (approx)

Step 2: Use the ideal gas law to find the number of moles of CH4.

n = PV / RT = (1.242 atm * 6.15 L) / (0.0821 L·atm/K·mol * 356.62 K) = 0.262 mol CH4 (approx)

Step 3: Use the balanced chemical equation for the combustion of methane to find the number of moles of H2O produced.

CH4 + 2O2 -> CO2 + 2H2O

From the balanced equation, we can see that 1 mol of CH4 produces 2 mol of H2O. Therefore, 0.262 mol CH4 will produce 0.262 * 2 = 0.524 mol H2O.

Step 4: Convert moles of H2O to grams.

The molar mass of H2O is approximately 18.015 g/mol. Therefore, 0.524 mol H2O * 18.015 g/mol = 9.44 g H2O (approx).

So, approximately 9.44 grams of H2O is formed.