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    • Thread Starter

    4 (NH3) + 5 (O2) => 4 (NO) + 6 (H2O)

    a) What volume of air is required by 1.00m3 of ammonia? (Take air to be 20% oxygen and 80% nitrogen)

    b) Assuming the reaction is complete, what volumes of what gases will be preset in the mixture leaving the catalyst for an input of 1.00m3 of ammonia? (Assume that the gas volumes are all measured at the temperature of the catalyst)

    Any help would be so much appreciated

    (Original post by kmdaccount)
    10cm3 of unknown hydrocarbon was sparked with 100cm3 of oxygen (an excess). When the resulting gases were cooled back to the original room temperature, they had a volume of 75cm3. Exposure of the gases to sodium hydroxide solution reduced the volume to 35cm3. Find the formula of the hydrocarbon.

    Any help would be much appreciated
    If these are all at RTP, remember that 1 mole of a gas occupies 24dm3 (24000cm3). So convert the volumes to moles.

    Remember that with these kinds of questions you need to take into account any remaining oxygen. The sodium hydroxide bit probably refers to water vapour being absorbed (not 100% sure of this).

    Have you done any basic working out and tried to solve this?
    • Thread Starter

    (Original post by y.u.mad.bro?)
    Have you done any basic working out and tried to solve this?
    Here is my working...

    1m3 = 1000000cm3
    106 / 24000 = 125/3 moles
    molar ratio of ammonia to oxygen:
    125/s moles of ammonia : 625/12 moles of oxygen
    635/12 moles multiplied by 24000 = 1250000cm3 O2 needed

    1250000/20 multiplied by 100 = 6250000cm3 air

    part a = 6.25m3 of air

    For part b I have worked out the volume of NO and H20 (due to the high temp)
    moles of NO = 125/3 moles
    125/3 moles multiplied by 24000 = 1000000cm3

    moles of H20 = 62.5
    62.5 multiplied by 24000 = 1500000cm3


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