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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 work used to solve this problem.

CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)

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Answer is: 4.45 grams of methane gas need to be combusted.
Balanced chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.
Ideal gas law: p·V = n·R·T.
p = 1.1 atm.
T = 301 K.
V(H₂O) = 12.5 L.
R = 0,08206 L·atm/mol·K.
n(H₂O) = 1.1 atm · 12.5 L ÷ 0,08206 L·atm/mol·K · 301 K.
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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