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    yassinjamal's Avatar
    yassinjamal Posts: 1, Reputation: 1
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    #1

    Nov 5, 2011, 08:22 PM
    Need help with a problem involving the calculation of ammonia gas
    In an industrial process, 400kg of ammonia gas, NH3, is oxidized in the presence of a catalyst to form nitrogen monoxide, NO, according to the following equation 4NH3 5o2=4NO 6H2O,
    calculate the number of moles of ammonia used in the reaction
    jcaron2's Avatar
    jcaron2 Posts: 986, Reputation: 204
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    #2

    Nov 5, 2011, 09:08 PM
    The formula that was provided for producing nitric oxide is interesting, but not necessary to answer the question. All they're really asking is "How many moles of NH3 are there in 400kg?"

    The first thing you need to figure out is this: What's the mass of 1 mole of NH3? Well, that's the very definition of the molecular weight - the mass of one mole of molecules. So what's the molecular weight of NH3? That's easy; just add up the atomic weights of the various atoms that comprise it. 1 mole of NH3 has 1 mole of N atoms and 3 moles of H atoms. You can look up the atomic weights of N and H on a periodic table. Add up the weights of each of the atoms, and you've got the molecular weight.

    Once you know the molecular weight (i.e. the mass of 1 mole of NH3 molecules), you just have to find out how many times that divides into 400kg. Remember, the molecular weight you calculate based on the periodic table is most likely going to be in grams. You'll need the molecular weight and the total mass of NH3 to be in the same units before you do the division, so you'll need to either convert 400kg into grams or convert the molecular weight into kg.

    If you get stuck on any of these steps, let us know where you're having trouble and one of us will provide more guidance. Otherwise, post back with the answer you get and we can check it for you to make sure it's right.

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