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4 April, 00:14

If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?

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  1. 4 April, 03:30
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    First, write the chemical reaction a nd balance it:

    NBr3 + NaOH - - - > N2 + NaBr + HOBr

    2 NBr3 + 3 NaOH - - - > N2 + 3 NaBr + 3 HOBr

    Then the theoretical proportion is 2 NBr3 : 3 NaOH

    The reactants are in the proportion 40 NBr : 48 NaOH

    From the theoretical proportion you can make [2 NBr3 / 3 NaOH] * 48 NaOH = 32 NaOH.

    Given that you have more than 32 moles of NaOH, this is the excess reactant.
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