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13 April, 20:54

Calculate the yield of NH3 in liters at STP, if you react 112 L of N2 with 134.4 L of H2, given the reaction: N2 + 3 H2 - -> 2 NH3 How many moles of N2 are given?

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  1. 13 April, 21:03
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    There will be produced 89.6 L of NH3

    There are 5 moles of N2 given

    Explanation:

    Step 1: Data given

    Volume of N2 = 112 L

    Volume of H2 = 134.4 L

    Step 2: The balanced equation

    N2 + 3 H2 → 2 NH3

    Step 3: Calculate moles of H2

    For 22.4 L we have 1 mol

    Number of moles = 134.4 / 22.4 = 6 moles

    Step 4: Calculate moles of N2

    Number of moles = 112 / 22.4 = 5 moles

    Step 5: Calculate limting reactant

    For 1 mol of N2 we need 3 mol of H2 to produce 2 moles of NH3

    The limiting reactant is H2. It will be completely be consumed (6 moles)

    N2 is in excess. There will react 6 moles / 3 = 2 moles

    There will remain 5 - 2 = 3 moles

    Step 6: Calculate moles of NH3

    For 1 mol of N2 we need 3 mol of H2 to produce 2 moles of NH3

    For 6 moles of H2 we'll have 4 moles of NH3

    Step 7: Calculate volume of NH3

    1 mol = 22.4 L

    4 moles = 89.6 L

    There will be produced 89.6 L of NH3
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