anonymous
  • anonymous
When I do double displacement for this formula (NH4)2CO3 + CaCl2 do I do it as the following? CaCO3 + (NH4)2Cl2 or (Ca)2CO3 + NH4Cl2
Chemistry
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
I tried this out and i got (NH4)2CO3 + CaCl2 --> NH4Cl + CaCO3
anonymous
  • anonymous
NH4 is a cation with a charge of +1. When it combines with the anion Cl which has a charge of -1, you don't need any coefficients to balance the formula out. Its just NH4Cl. Similarly, Ca has a charge of +2 and CO3 has a charge of -2 so they don't need any coefficients either.
anonymous
  • anonymous
But if I needed to write out it's ionic equation format do I need to have it like (NH4)2 or can I just drop the ()2 and leave NH4?

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anonymous
  • anonymous
if it's in the ionic equation, you need to make sure that you have the same number of NH4 ions on both sides of the equation. In the reagents side, you have 2 ions of NH4 in the molecule, so you need to have 2 ions of NH4 in the products side. the format that you should write it in is 2 NH4+, not (NH4)2
anonymous
  • anonymous
So it should look like this? (NH4)2CO3 + CaCl2 →CaCO3+(NH4)_2Cl Ionic equation CaCO_3+(NH_4)_2Cl→ 〖Ca〗^(2+) CO_3^(2-)+2〖NH〗_4^++〖Cl〗^- Net Ionic Equation CaCO_3+(NH_4)_2Cl→ (NH_4)_2Cl
anonymous
  • anonymous
Sorry missed a couple things. (NH4)2CO3 + CaCl2 →CaCO3+(NH4)2Cl Ionic equation CaCO3+(NH4)_2Cl→ 〖Ca〗^(2+)+CO_3^(2-)+2〖NH〗_4^++〖Cl〗^- Net Ionic Equation CaCO3+(NH4)2Cl→ (NH4)2Cl
anonymous
  • anonymous
This is the complete equation: \[(NH _{4})_{2}CO _{3} + CaCl _{2} \rightarrow CaCO _{3} + 2 [NH _{4}Cl]\] This is the complete ionic equation: \[2[NH _{4}^{+}] + CO _{3}^{-} + Ca ^{2+} + 2[Cl ^{-}] \rightarrow 2[NH_{4}^{+}] + CaCO _{3} + 2Cl ^{-}\] Using a solubility table, we can see that CaCO3 is inosoluble so it doesn't dissociate. What is the net ionic equation then?
anonymous
  • anonymous
So sorry CaCO3 is the net ionic equation. So is the rest correct?
anonymous
  • anonymous
For the net ionic equation, cross out the parts that are exactly the same on both sides. So, in this case, that's the NH4 ion and the Cl ion. Eliminate those. That leaves you with Ca2+ + CO3 2- --> CaCO3 as the net ionic equation.
anonymous
  • anonymous
Right. I get that. I forgot I have to keep the insoluble for the net ionic. how did the Ionic equation look. Did it look correct.
anonymous
  • anonymous
your ionic equation is almost correct, but remember to keep the order of the equation the same as your original equation. you put the products on the side of the reagents, when it should be the other way around. Also, NH4Cl is soluble, so it will dissociate. so you should have the ions separate.
anonymous
  • anonymous
and when you write NH4Cl, the 2 goes in front of the entire compound, not after the NH4. There are 2 moles of the compound NH4Cl, not 1 mole of (NH4)2Cl
anonymous
  • anonymous
But wouldn't I have to do double displacement for this formula to make this work or is that an optional Idea. In the book it shows that you have to do double displacement. Also sorry products and reagents???
anonymous
  • anonymous
So switch it like this? 2NH4Cl+CaCO3
anonymous
  • anonymous
Products = the products of the reaction. those are on the right side of the complete equation that i posted. Reagents = the compounds that react together to form the products. They're on the left side of the complete equation. The double displacement still happens. If you look at the complete equation, the compounds each switch "partners" - The NH4 starts off in a compound with the CO3, and finishes in a compound with the Cl. When you write out the ionic equation, you are dissociating any compounds that are soluble. It doesn't matter what kind of reaction youre doing (single displacement, double displacement, etc); if the compound is soluble, dissociate it into the separate ions. Then you're pretty much doing algebra with the ions and insoluble compounds. Some may cancel each other out. That's why the net ionic equation may look like a formation reaction when it's really not.
anonymous
  • anonymous
Ok but what if I have two solutions that are soluble how am I supposed to find the Net Ionic Equation. NaNO3 + KCl Right off the bat according to my chart all of these are soluble.
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anonymous
  • anonymous
if every compound in an equation is soluble, then there really is no net ionic equation. every ion involved is a spectator ion. The correct answer for a question like that is "no net ionic equation"
anonymous
  • anonymous
So even if I went through the process of double displacement and finding the ionic equation that would be the answer?
anonymous
  • anonymous
Yup. You can find an ionic equation, just no net ionic equation.
anonymous
  • anonymous
Ok thank you very much.

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