Question

How many liters of water vapor can be produced if 102 grams of methane gas (CH4) are combusted at 315 K and 1.2 atm? Show all of the work used to solve this problem. CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)

Answer 1

Answer:
V = 274.77 L

Solution:

First find out the moles of CH₄ as,

Moles = Mass / M/mass

Moles = 102 g / 16 g.mol⁻¹

Moles = 6.375 mol

Now, according to equation,

1 mole CH₄ produced = 2 moles of H₂O vapours
So,
6.375 mol of CH₄ will produce = X moles of H₂O vapours

Solving for X,
X = (6.375 mol × 2 mol) ÷ 1 mol

X = 12.75 mol of H₂O

Now, find out volume of H₂O as,

P V = n R T
Or,
V = n R T / P

Putting values,

V = (12.75 mol × 0.0821 atm.L.mol⁻¹.K⁻¹ × 315 K) ÷ 1.2 atm

V = 274.77 L