10 pnts best answer!! writing balanced equations for each of the reactions formed.?

Write balanced chemical equations for each of the reactions performed. Be sure to show the state of the reactant and product. if no reaction, write "no reaction":

1) sodium hydroxide and copper(2) sulfate

2) sodium chloride and potassium nitrate

3) sodium hydrogen carbonate and hydrochloric acid

4) barium chloride and sodium sulfate

5) hydrochloric acid and sodium hydroxide

6) zinc nitrate and copper(2) sulfate

7) sodium hydroxide and iron(3) chloride

8) sodium sulfite and hydrochloric acid

9) ammonium chloride and copper sulfate

10) lead(2) nitrate and potassium iodide

please show how to do these and the final answers. thanks alot!!!

10 pnts best answer!! writing balanced equations for each of the reactions formed.?sunshine

1. 2NaOH(aq) + CuSO4(aq) --%26gt; Na2SO4(aq) + Cu(OH)2(s)

This is one example of a double-displacement reaction. When you write a chemical equation, make sure you know some common cations and anions as well as the rules on writing chemical compounds.

Sodium hydroxide is written as NaOH. The ions are Na+ and OH-. Interchange the charges of Na and OH respectively and you get NaOH. Remember that the charges stand as the SUBSCRIPTS of the atoms involved.

From your solubility rules, Cu(OH)2 is insoluble. Look it up. Here's the website: http://www.ausetute.com.au/solrules.html Read it so you'll get an idea.. :D

NaCl(aq) + KNO3(aq) --%26gt; NaNO3(aq) + NaCl(aq)

This is a double-displacement reaction. Just interchange positions of the cations. Based on your solubility rules, no precipitate will be formed because all are soluble.

NaHCO3(s) + HCl(aq) --%26gt; NaCl(aq) + CO2(g) + H2O(l)

You could've written the reaction as

NaHCO3 + HCl --%26gt; NaCl + H2CO3

but H2CO3 is quite unstable and will tend to break up into carbon dioxide and water. (Don't worry, it's still a double-displacement reaction... ;) )

Remember that the formula for the anion hydrogen carbonate is HCO3-. It is also called bicarbonate.

When I wrote the products for this equation, I first started out with H2CO3 and NaCl respectively. I simply interchanged the positions of the cations involved... :)

For that H2CO3, I thought to myself that I had the H+ and the CO3[2-] ions involved. I simply interchanged the charges on both to act as the subscripts. It's also known as carbonic acid, based from your rules on chemical nomenclature.

BaCl2(aq) + Na2SO4(aq) --%26gt; BaSO4(s) + 2NaCl(aq)

Again, based on your solubility rules, sulfates of barium are insoluble and they precipitate out of solution, hence I wrote the state symbol (s). For the way I wrote out the chemical formulas of the reactants, just remember some of your common cations and anions. Interchange the charges of your cations and anions so you can come up with your compound. It would be able to have a list of common cations and anions at hand.

HCl(aq) + NaOH(aq) --%26gt; NaCl(aq) + H2O(l)

This is a neutralization reaction. A neutralization reaction involves an acid and base, and they give of a salt and water as products.

Zn(NO3)2(aq) + CuSO4(aq) --%26gt; ZnSO4(aq)+ Cu(NO3)2(aq)

Another double-displacement reaction here. Both are soluble with each other, so no precipitates.

Why is the formula for zinc nitrate Zn(NO3)2?

If you would note, the charge of Zn is 2+ and that of nitrate is -1. Interchanging the charges gives us ZnNO32.

Oops. We don't want that to happen. This would be an entirely different formula. To compensate for that, we need to write parentheses before and after the nitrate ion.

Zn(NO3)2

Now, we can say that there are two nitrate ions for every zinc ion.

3NaOH(aq) + FeCl3(aq) --%26gt; 3NaCl(aq) + Fe(OH)3(s)

We had Fe3+ as the cation for FeCl3. We know, because when we interchanged the charges, we're able to deduce that Cl- had a charge of -1 and Fe had +3. Now, when we wrote the formula for Fe(OH)3, we used the +3 charge of Fe.

Based from the solubility rules, the hydroxide of Fe is insoluble so I wrote its state symbol as (s), to tell that that hydroxide precipitated out of reaction.

Na2SO3(aq) + 2HCl(aq) --%26gt; 2NaCl(aq) + H2SO3(aq)

Remember that the sulfite anion is written as SO3[2-]. Be able to distinguish the sulfate and sulfite anions. Sulfate has 4 oxygen atoms while sulfite has 3.

So, the acid involved should be written as sulfurous acid because the anion involved is the sulfite ion. If that were the sulfate, it would be sulfuric acid. Remember to use -ous for the one with the lower number of oxygen atoms and -ic for the one with the higher number of oxygen atoms. (Use this only if the ion in question has variants in the number of oxygen atoms. If otherwise, stick with writing the ternary acid as __ic acid).

NH4Cl(aq) + CuSO4(aq) --%26gt; (NH4)2SO4(aq) + CuCl2(s)

I presume the copper in copper sulfate must be Cu2+. CuCl2 precipitates out of reaction.

Pb(NO3)2 + KI --%26gt; No reaction

K is more reactive than Pb. This is from your activity series. Look it up. The element being displaced should be lower than the one doing the displacing. In this case, Pb is lower than K in the activity series. So Pb cannot displace K.

:)