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Chemistry

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If 10g of NO react with 20g of O2, what is the maximum amount of NO2 that can be produced? How do I know exactly which is limiting and which is excess, is there a calculation that shows this?

  • Chemistry - ,

    In order to find the limiting and/ or excess reactant, you have to first convert grams to moles, then once you have that you did the #s by the smallest mole #. Once you have that, you compare that to the coefficients of the balanced equation.

  • Chemistry--Help!! - ,

    This was my answer, but it was wrong:

    2NO + O2 ----> 2NO2

    Molar Mass NO2 = 46g/mol
    Molar mass NO = 30g/mol
    Molar Mass O2 = 32g/mol

    0.333 mols NO x 46g NO2/1mol NO = 15.3g NO2

    mols NO = 10g NO/30g/1mol NO = 0.333 mols NO
    mols O2 = 20g O2/32g O2/1mol O2 = 0.625 mols O2

    Where did I mess up?? And how do I figure out what is limiting reactant and what is excess?

  • Chemistry---DrBob222 - ,

    Am i solving this correctly, according to my answer/posts above??

  • Chemistry - ,

    Ok, so you just do .333/.333 =1 (NO)
    and .625/.333= 1.87 (O2)

    So, then you go back to your equation, and you see how 1 < 2(# of moles of NO in the equation), that means it is the limiting reactant

    Then you do the same thing with 1.87, and 1.87> 1( # of mole of O2), so that tells you it is the excess reactant. Make sense?

  • Chemistry - ,

    I guess so, thanks "K"

  • Chemistry - ,

    Tmofey, you didn't do anything wrong in your chemistry. I think the problem is math. I see 20g and 30g as having 2 significant figures so you can't report more than 2 in your answer. If you are keying this into a database, as I suspect, those things are notoriously unforgiving for the number of s.f. So your answer is 15.3 g NO2, you are allowed only 2 s.f., round the answer to 15 g NO2 produced. I'll bet the database accepts that answer.

  • Chemistry - ,

    I will do that, thanks for your help once again DrBob222.

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