Question

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

Answer 1

V = 33,5 L

Step-by-step 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 2

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.