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just olya [345]
3 years ago
12

How many grams of methane gas (CH4) need to be combusted to produce 12.5 L water vapor at 301 K and 1.1 atm? Show all of the wor

k used to solve this problem.
CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)
Chemistry
1 answer:
tangare [24]3 years ago
4 0
Answer is: 4.45 grams of methane gas <span>need to be combusted</span>.
Balanced chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.
Ideal gas law: p·V = n·R·T.<span>
p = 1.1 atm.
T = 301 K.
V(H</span>₂O) <span>= 12.5 L.
R = 0,08206 L·atm/mol·K.
</span>n(H₂O) = <span>1.1 atm · 12.5 L ÷ 0,08206 L·atm/mol·K · 301 K.
</span>n(H₂O) = 0.556 mol.
From chemical reaction: n(H₂O) : n(CH₄) = 2 : 1.
n(CH₄) = 0.556 mol ÷ 2 = 0.278 mol.
m(CH₄) = 0.278 mol · 16 g/mol.
m(CH₄) = 4.448 g.
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