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# how many liters of O2 (g) are needed to react completely w/ 56.0 L of CH4 (g) at STP to produce CO2 (g) and?

H2O (g)? GIVEN: CH4 (g) + 202(g)>>>CO2(g) + H2O (g)

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- Anonymous1 decade agoFavorite Answer
the balanced equation is

CH4 + 2H2O>> CO2 + 8H+ + 8e-

(O2 + 4H+ + 4e->> 2H2O) x 2

CH4 + 2O2 >> CO2 + 2H2O

STP >> p= 1 atm and T = 273 K

pV = nRT

n = pV / RT = 1 x 56.0 / 0.0821 x 273 = 2.5

the ratio between CH4 and O2 is 1 : 2 so we need 5.0 moles of O2

V = nRT / p = 5 x 0.0821 x 273 / 1 = 112 L of O2

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