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I am supposed to write the precipitation reaction for

$\ce{(NH4)2C2O4 + AgNO3 ->[\ce{H2O}]}$

Is my equation correct? I left out the Spector ions. $$\ce{4(NH4) + 2 C2 +O4\rightarrow 4(NH4)2CO3}$$

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  • $\begingroup$ How can your equation be correct if the number of each element on both sides are not the same? $\endgroup$
    – LDC3
    Apr 4 '14 at 3:32
  • $\begingroup$ Good morning! I took the liberty to add some tags and edited the title and the body of your question. I hope you are OK with the changes. I find it a tad more understandable this way, understood it and tried to give an answer that might help you to tackle questions like this. $\endgroup$ Apr 4 '14 at 6:00
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Your current solution is wrong and you don't seem to be familiar with the concepts involved.

Let's walk through this together and change that!

  1. You take solid ammonium oxalate $\ce{(NH4)2C2O4}$ and silver nitrate $\ce{AgNO3}$ and dissolve both salts in water.

  2. The salts dissociate into their ions. Let's do that separately for both salts first.

    $$\ce{ (NH4)2C2O4 ->[\ce{H2O}] 2 NH4+_{(aq)} + C2O4^{2-}_{(aq)}}$$

    $$\ce{AgNO3 ->[\ce{H2O}] Ag+_{(aq)} + NO3^{-}_{(aq)}}$$

    Since you are dissolving both salts in the the same beaker with water, your solution now contains all the ions of both equations.

    Forget about the spectator ion fanciness.

    It is much more important to identify the chemical units (= the ions from which the salts are composed). Leave these ions intact and don't rip them apart into weird combinations of elements.

  3. Precipitation will occur when the combination of these ions (chose a cation and an anion) yields a salt with a very low solubility in water. There are only two cations and anions here, so the number of possible combinations isn't too big. Look up the solubilities of the possible products to figure out which one will precipitate.

  4. Do not forget to balance the reaction!

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