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Finding limiting reactant

  1. Mar 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Procedure
    1. Add 25 mL of HCL solution to each flask.
    2. Weigh out 0.15g, 0.3g, and 0.6g of Mg ribbon and place each sample into its own balloon




    2. Relevant equations

    Mg + 2HCl --> H2 + MgCl2

    3. The attempt at a solution

    show the calculations determining the limiting reactant for each reaction.

    Help on how to get the limiting reactant!!

    And using the limiting reactant as the starting amount, determine the amount of hydrogen gas that was produced per reaction flask:
     
  2. jcsd
  3. Mar 6, 2008 #2
    I think you need to know the molarity of HCl.
     
  4. Mar 6, 2008 #3
    oh my bad 1.0moles HCL/1L or 0.1 moles HCL/100mL solution
     
  5. Mar 6, 2008 #4
    Since you know the molarity of HCl to be 1M, you can compute how many moles of HCl you have at 25mL. The formula Molarity = mole/volume in Liter, or mole = Molarity x Volume in liter.
    Further, based on your equation, you know you need a 2 HCL to 1 Mg ratio.
    So just figure out how many moles of Mg you have at .15g, .3 and .6
    From that you can see your limiting reactant.
     
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