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

V = 33,5 L

Explanation:

For the reaction of combustion of methane:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

12,0g of methane are:

12,0g × (1mol / 16,04g) = 0,748 moles of CH₄

Based on the reaction, 1 mole of CH₄ produce 2 moles of H₂O. Thus, 0,748 moles of methane produce:

0,748 moles CH₄ × (2mol H₂O / 1mol CH₄) = 1.496 moles of H₂O

Using gas law it is possible to know the volume, in liters of water at STP, thus:

V = nRT / P

Where n are moles (1.496 moles of H₂O), R is gas constant (0.082atmL/molK), T is temperature (273.15K at STP) and P is pressure (1atm at STP). Replacing:

V = 1.496moles × 0.082atmL/molK× 273.15K / 1 atm

V = 33,5 L

I hope it helps!

answer is d because it is trust me