How many liters of water vapor, at stp, are produced by the combustion of 12.0 g of methane?

Answer is: 33,6 liters of water vapor.

Chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.

m (CH₄) = 12 g.

n (CH₄) = m (CH₄) : M (CH₄)

n (CH₄) = 12 g : 16 g/mol = 0,75 mol

from reaction: n (CH₄) : n (H₂O) = 1 : 2

n (H₂O) = 1,5 mol.

V (H₂O) = n (H₂O) · Vm

V (H₂O) = 1,5 mol · 22,4 mol/dm³

V (H₂O) = 33,6 dm³.