How many grams of methane gas (CH4) need to be combusted to produce 12.5 L water vapor at 301 K and 1.1 atm? Show all of the work used to solve this problem.

CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)

Answer is: 4.45 grams of methane gas need to be combusted.

Balanced chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.

Ideal gas law: p·V = n·R·T.

p = 1.1 atm.

T = 301 K.

V (H ₂O) = 12.5 L.

R = 0,08206 L·atm/mol·K.

n (H₂O) = 1.1 atm · 12.5 L : 0,08206 L·atm/mol·K · 301 K.

n (H₂O) = 0.556 mol.

From chemical reaction: n (H₂O) : n (CH₄) = 2 : 1.

n (CH₄) = 0.556 mol : 2 = 0.278 mol.

m (CH₄) = 0.278 mol · 16 g/mol.

m (CH₄) = 4.448 g.