Consider the following chemical reaction 2HCL+ Ca(OH)2 CaCl2+2H2O
You have mixed 30.00ml of 0.150 M HCl solution with 20.00ml of 0.100m Ca(OH)2 solution what is the molarity of the HCl, Ca( OH)2 and CaCl2 after the reaction has stopped

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  1. This is a limiting reagent (LR) problem with a twist or two.
    mols HCl = M x L = about 0.0045
    mols Ca(OH)2 = M x L = about 0.002

    mols CaCl2 formed if we start with 0.0045 mols HCl and all of the Ca(OH)2 we need is
    0.0045 mols HCl x (1 mol CaCl2/2 mols HCl) = 0.0045 x (1/2) = about 0.00225.

    mols CaCl2 formed if we started with 0.002 mols Ca(OH)2 and all of the HCl we needed is
    0.002 x (1 mol CaCl2/1 mol Ca(OH)2) = 0.002.

    You have two different values for mols CaCl2 which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
    So all of the Ca(OH)2 will be used and you will form 0.002 mol CaCl2. Thus
    [Ca(OH)2] = 0
    [CaCl2] = 0.002 mols/0.05L = ?

    For HCl we must determine how much of it has been used and how much is left.
    mols used = 0.002 mol Ca(OH)2 x (2 mol HCl/1 mol Ca(OH)2) = 0.002 x 2 = 0.004.
    We started with 0.0045 so we have left the difference or 0.0005 mols and that divided by the volume (in L) is the M.
    By the way, you need to find and use the arrow key. You can make arrows either --> or ==>. And note to be careful with the M and m since they mean different things.

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