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Net ionic equations (NaOH; HCl; H2O)

  1. Oct 4, 2006 #1

    I completed a Grade 12 experiment and now I am doing some calculations about it but I am confused.

    Here is the question: "Write the net ionic equation for each reaction, and note the value of (triangle sign) H for each reaction. e.g. H+(aq) + OH-(aq) --> H2O(l)
    (triangle)H2 = -45 kJ/mol HaOH

    I had three reactions.
    1. Heat of dissolution of NaOH; 200 mL of labelled water + 5.5 g of crystal NaOH

    2. Heat of reaction between aqueous NaOH and aqueous HCl; 100 mL of 1.0 mol/L HCl + 100 mL of 1.0 mol/L NaOH

    3. Heat of reaction between solid NaOH and aqueous HCl; 200 mL of 1.0 mol/L HCl + 5.5g of crystal NaOH

    These were the instructions also given: "Since all the solutions are dilute, then the density of each solution can be assumed to be 1.00 g/mL. So, 100 mL has a mass of 100 g."

    So for the first reaction, would the net ionic equation be something like this:
    1. 200 g H2O(l) + 5.5 g NaOH(s) --> I don't know what the product would be?
    2. 100 g HCl(aq) + 100 g NaOH(aq) --> NaCl + H2O?
    3. 200 g of HCl(aq) + 5.5 g NaOH(s) --> NaCl + H2O?

    Any help would really be appreciated!:confused:
  2. jcsd
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