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18 February, 00:52

In one experiment, 0.893 mol of NO is mixed with 0.519 mol of O2. Determine which of the two reactants is the limiting reactant. Calculate also the number of moles of NO2 produced.

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  1. 18 February, 04:22
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    NO is the limiting reactant.

    0.893 moles of NO2 will be produced

    Explanation:

    Step 1: Data given

    Number of moles NO = 0.893 moles

    Number of moles O2 = 0.519 moles

    Step 2: The balanced equation

    2NO + O2 → 2NO2

    Step 3: Calculate the limiting reactant

    For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

    NO is the limiting reactant. It will completely be consumed. (0.893 moles)

    O2 is in excess. There will react 0.893/2 = 0.4465 moles.

    There will remain 0.519 - 0.4465 = 0.0725 moles O2

    Step 4: Calculate moles NO2

    For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

    For 0.893 moles NO we'll have 0.893 moles NO2

    NO is the limiting reactant.

    0.893 moles of NO2 will be produced
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