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14 October, 02:50

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO: 4NH3 (g) 5O2 (g) - -> 4NO (g) 6H2O (g) In a certain experiment 2.00 g of NH3 reacts with 2.50 g of O2. Which is the limiting reactant?

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  1. 14 October, 03:27
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    O2 is the limiting reactant.

    Explanation:

    Step 1: Data given

    Mass of NH3 = 2.00 grams

    Mass of O2 = 2.50 grams

    Molar mass NH3 = 17.03 g/mol

    Molar mass O2 = 32 g/mol

    Step 2: The balanced equation

    4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (g)

    Step 3: calculate moles NH3

    Moles NH3 = mass NH3 / molar mass NH3

    Moles NH3 = 2.00 grams / 17.03 g/mol

    Moles NH3 = 0.117 moles

    Step 4: Calculate moles O2

    Moles O2 = mass / molar mass O2

    Moles O2 = 2.50 grams / 32 g/mol

    Moles O2 = 0.0781 moles

    Step 5: Calculate the limiting reactant

    For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

    O2 is the limiting reactant. It will completely be consumed. (0.0781 moles). NH3 is in excess. There will react 4/5 * 0.0781 moles = 0.0625 moles

    There will remain 0.117 - 0.0625 = 0.0545 moles NH3

    O2 is the limiting reactant.
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