March 27, 2017

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My teacher hasn't really talked about types of reactions; he's just given us a list of net ionic equations involving water and different elements including Lithium, Sodium, Calcium, Magnesium, Lanthanum, and Potassium to memorize. However, his test questions don't involve the reactants/net ionic equations he's given us, so when I see "Show a net ionic equation for the reaction between Water and Strontium metal" I don't know what to do.

  • Chemistry - ,

    Elements in Group IIA of the periodic table can replace one of the hydrogen atoms in H2O to produce a hydroxide and H2 gas. Example:
    Ca(s) + H2O ---> Ca(OH)2 + H2
    Ca(s) + 2H2O ---> Ca(OH)2 + H2 (balanced)
    Other elements in the same column as Ca react in a similar way.
    Elements in Group IA of the periodic table can replace one of the hydrogen atoms in H2O, also, to produce a hydroxide and H2 gas. Example:
    Li + H2O ---> LiOH + H2
    2Li + H2O ---> 2LiOH + H2 (balanced)
    Other metals in the same column as Li react in a similar way.
    La reacts in a similar way but the products are La(OH)3 and H2. Try writing and balancing the chemical equation.

  • Chemistry - ,

    Has your teacher given you the molecular equation. For Na, that is
    Na + H2O ==> NaOH + H2 and balanced it will be
    2Na + 2H2O ==> 2NaOH + H2

    Now what you want to do is to show the form as follows. To simplify things, I will not worry about the balancing coefficients.
    2Na(s) + 2H2O(l) ==> 2NaOH(aq) + H2(g)
    aq = aqueous
    Now you want to take the molecular equation and make that into an ionic one.
    solids don't change.
    liquids don't change
    aq (separate into ions).
    gases don't change.
    Na(s) + H2O(l) ==> Na^+(aq) + OH^-(aq) + H2(g)
    Sr metal, Ca metal, K metal, any of the metals that will react with water will do so to produce the corresponding base (the OH part) plus H2(g).
    All of the ion part will be analogous. The final step in making a NET ionic equation is to subtract (cancel) those items that are common to both sides. In the above equation, the final ionic equation IS the net ionic equation because nothing cancels.
    I will do the Pb(NO3)2 and LiCl, again but I will use the balanced equation.
    The molecular equation is
    Pb(NO3)2 + 2LiCl ==> PbCl2 + 2LiNO3
    You had aqueous solution so I will add that now.
    Pb(NO3)2(aq) + 2LiCl(aq)==>PbCl2(s) + 2LiNO3(aq)

    Now change these to ions.
    Pb(NO3)2 is soluble; therefore, it becomes the ions, LiCl ditto. PbCl2 is insoluble and stays as is. LiCl is soluble and changes to ions.
    Pb^+2(aq) + 2NO3^-(aq) + 2Li^+(aq) + 2Cl^-(aq) ==> PbCl2(s) + 2Li^+(aq) + 2NO3^-(aq).

    That last is the molecular equation changed to the ionic equation. Now to make the net ionic equation, we cancel those items common to both sides. In this case there is no Pb^+2 on the right so we keep it on the left. Notice there are 2NO3^- on the left and right; therefore, cancel them. They will not show up in the final equation. There are 2Li^+ on both sides so cancel them. They will not show up either. What's left?
    Pb^+2(aq) + 2Cl^-(aq) ==> PbCl2(s)
    This is the net ionic equation. Sorry it took me so long but this isn't easy to type.

  • Chemistry - ,

    I appreciate your patience! Thank you.

  • Chemistry - ,


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