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Percent yield/excess reactant problem

  1. Nov 18, 2017 #1
    1. The problem statement, all variables and given/known data

    N2 reacts with H2 to form NH3

    assume equal moles of N2 and H2 in container, and react to form NH3 with 100% yield. there are 2.0 moles of excess reactants left over. How many moles of each reactant was originally present?

    2. Relevant equations
    AY/TY = % yield

    3. The attempt at a solution
    I know that the answer is 3 moles after much trial and error. Cause if i had started with 3 moles of each reactant, then i used 1 mole of N2. Which means 2 moles in excess. How could i have shorten this problem without doing trial error?
  2. jcsd
  3. Nov 19, 2017 #2

    I like Serena

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    Homework Helper

    Hi Lori! :oldsmile:

    We can set up equations, or we can do 'smart' trial and error.

    With equations
    We have 2 unknowns: the initial amount of reactants, and the amount of NH3 formed.
    Let's call them x and y.
    Then the equation becomes: x N2 + x H2 → y NH3 + 2 N2
    Can we make a set of equations from that and solve it?

    With 'smart' trial and error
    Let's first balance the equation:
    ? N2 + ? H2 → ? NH3 + ? N2
    Can we find the question marks such that the equation is balanced?
    After that we need to figure out what to do to ensure we're left with 2 moles of initial reactant.
    Last edited: Nov 19, 2017
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