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15 March, 02:43

If you have 20L of nitrogen, and 25L of hydrogen, how many molecules of NH3 can you make? How much excess reactant is left over?

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  1. 15 March, 04:08
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    The balanced equation for formation of NH₃ is as follows;

    N₂ + 3H₂ - - > 2NH₃

    Stoichiometry of N₂ to H₂ is 1:3

    when gases react, stoichiometry of reactants applies to volumes as well.

    Because 1 mol of any substance occupies 22.4 L at STP.

    Therefore stoichiometry of volumes of N₂ to H₂ is 1:3

    If N₂ is the limiting reactant,

    1 L of N₂ reacts with 3 L of H₂

    Therefore 20 L of N₂ should react with - 3/1 x 20 = 60 L

    Only 25 L of H₂ is present therefore H₂ is the limiting reactant

    ratio of H₂ to NH₃ is 3:2

    if 3 L of H₂ forms - 2 L of NH₃

    then 25 L of H₂ forms - 2/3 x 25 L = 16.67 L of ammonia

    if 22.4 L of NH₃ contains 1 mol of NH₃

    then 16.67 L of NH₃ consists of - 1/22.4 x 16.67 = 0.74 mol oh NH₃

    1 mol of NH₃ is made of 6.022 x 10²³ NH₃ molecules

    Therefore 0.74 mol of NH₃ made of 6.022 x 10²³ molecules/mol x 0.74 mol

    number of NH₃ molecules formed - 4.45 x 10²³ molecules of NH₃

    N₂ is in excess

    25 L of H₂ reacts with - 25/3 = 8.33 L

    however 20 L present initially

    Excess reactant left - 20 - 8.33 = 11.67 L of N₂ remaining
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