products of NaNO2 and HCl?

Despite the fact that there are four other answers so far, let me set this straight. On the one hand we have the group that says the products are NaCl and HNO2 as if there are actually molecules of NaCl formed, which, of course, there aren't. And then we have the lone hold out, Daniel, who says there won't be a reaction. The truth lies somewhere in between, but Daniel is closest.

Just because you mix two chemicals doesn't mean that you will get a reaction, and it is sometimes tricky reporting the formula of a compound when there is really no compound formed. Let us first consider what it is we're mixing. Since you didn't specify the state symbols with your formulas, we'll assume that HCl is in aqueous solution. Both HCl gas and solid sodium nitrite are quite soluble in water, so you will begin with a mix of sodium ions, nitrite ions, hydrogen ions and chloride ions just as Daniel has suggested. (Good job so far Daniel)

But here is where the rest of the crowd gets a point or two out of 10, because they got a partial right answer, but they don't know why they got the right answer. And this is where Daniel drops the ball. Nitrite ion, NO2^- and hydrogen ions will combine to make undissociated nitrous acid, HNO2 (or it is sometimes written as HONO.) Nitrous acid is a weak acid and only exists in solution, and it can be produced by the combination of nitrite ion and hydrogen ion. But since it is a weak acid, we represent its existence using an equilibrium expression because H+, NO2^- ions are present along with undissociated HNO2.

HNO2(aq) <==> H+ + NO2^- ...... Ka = 4.0 x 10^-4

So technically we could write the reaction as

NaNO2(s) + HCl(aq) <==> NaCl(aq) + HNO2(aq)

or better yet as

Na+ + NO2^- + H+ + Cl- <==> Na+ + Cl- + HNO2(aq)

but notice that Na+ and Cl- ions are the same on each side, so we write the net ionic equation

NO2^- + H+ <==> HNO2(aq) ...... Keq = 2.5x10^3

Rashmika suggests that NO2 gas is produced. A good idea, since HNO2 is unstable, but poor in execution. He hasn't taken into account the changes in oxidation sate, has left out a crucial product, and has not balanced the equations.

Here is the next step in HNO2's journey. HNO2 undergoes disproportionation (A redox reaction where the same species is both oxidized and reduced.) and forms NO2 and NO. In HNO2 nitrogen has an oxidation state of +3. In NO2 gas nitrogen has an oxidation state of +4, and in NO, nitrogen has an oxidation state of +2. NO2 gas actually reacts with water to make more HNO2 as well as HNO3. More disproportionation where nitrogen in the +4 oxidation state makes nitrogen in the +3 oxidation state, HNO2, and nitrogen in the +5 oxidation state, HNO3. When both reactions are taken together we have the overall reaction shown below.

2(2HNO2 --> NO2 + NO + H2O)

2NO2 + H2O --> HNO3 + HNO2

-----------------------------------------------

3HNO2 --> HNO3 + 2NO + H2O

Therefore, your overall reaction could be written as ....

3NaNO2(aq) + 3HCl(aq) --> 3NaCl(aq) + HNO3(aq) + 2NO(g) + H2O(l)

Nano2 Hcl

NaNO2 + HCl >< NaCl + HNO2

Since all of these compounds are soluble there is no net reaction, so there are no new products. All you will have is Na+, Cl -, H+, and NO2- ions floating in the solution.

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RE:

products of NaNO2 and HCl?

Can you please give the products of the reaction between NaNO2 and HCl? And also please give the balanced chemical equation! :) THANK YOU!

Na is in column 1 of the chart, so it has 1 + charge. Thus, NO2 must have 1 - charge. Cl is in column 7, so it must have 1 - charge. H, in this case (it varies) will have 1 +. That's an even trade, so you'll get salt (NaCl) and NHO2.

you will surely get hno2 (used in dizo preparation)...but as it readily decomposes to form no2 and NO, thus it is usually prepared in-situ...no2 liberation was observed as temp was not low enough to obtain stable hno2...further no2 my react with h2o (if present). so, if temp is low (around 273k) hno2 + nacl.....if temp is high...you will get no2 and NO

NaCl and HNO2