What is the total pressure when 1.00 L of methane (CH4) at STP and 4.00 L of oxy

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Explanation / Answer

22.4 L of any gas at STP = 1 mole => 1 L of any gas at STP = 1/22.4 mole = 0.0446 moles hence no of moles of gas = no. of moles of oxygen + no. of moles of methane = 4*0.0446 + 1*0.0446 = 0.223 moles total pressure = 0.223*0.08241*298/12 = 0.456 atm CH4 + 2(O2) ==> CO2 + 2(H2O) CH4 consumed = 0.0446 moles (given) O2 consumed = 2*0.0446 moles => O2 left = initially present - consumes = 4*0.0446 - 2*0.0446 = 0.0892 moles CO2 formed = 0.0446 moles H2O formed = 2*0.0446 moles = 0.0892 moles mole fraction of CO2 = 0.0446/(0.0446 + 0.0892 + 0.0892) = 1/5 = 0.2 mass fraction of CO2 = 0.0446*44/(0.0446*44 + 0.0892*32 + 0.0892*18) = 0.305 partail pressure of CO2 = 0.0446*0.08241*298/12 = 0.09127 atm

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