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Ammonia and oxygen react to form nitrogen and water.

4NH3(g)+3 O2(g)-2N2(g)+6H2O(g)

a.How many grams of O2 are needed to react with 8.00 mol NH3?

8.00 mol NH3 x42.094/4 mol NH3 x32.00 O2/1 mol O2=269gO2

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2 answers

  1. I don't know where the 42.094 came from.

    8.00 mole NH3.
    Convert to moles oxygen.
    8 x (3 moles O2/4 moles NH3) = 6.00 moles oxygen.

    Now convert 6.00 moles oxygen to grams. g = moles x molar mass.

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  2. 96 gm.

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