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)
1 answer:
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
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