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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: RayFlacco on December 11, 2016, 03:45:18 PM

Title: Reaction between ammounium chloride and sodium hydroxide
Post by: RayFlacco on December 11, 2016, 03:45:18 PM: Reaction is:

NH4Cl(g) + NaOH(aq) --> NaCl(aq) + NH3(g) + H2O(l)

Next question is to determine the amount of moles of NH4+, OH- og NH3 before and after the reaction aswell as the concentrations after the reaction. Can anyone help me with steps?
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: AWK on December 11, 2016, 03:51:30 PM: Some number data are missing.
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: RayFlacco on December 11, 2016, 03:58:02 PM: Oh that's on my part. What I know is that I used 1g ammonium chloride and 10ml 1M sodium hydroxide
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: AWK on December 11, 2016, 04:03:56 PM: Now you can calculate the stoichiometry of reaction.
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: RayFlacco on December 11, 2016, 04:11:05 PM: Yeah that's what I'm struggling with. Do you know somewhere I can read the steps or something?
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: AWK on December 11, 2016, 04:45:03 PM: Stoichiometry of reaction you can find in any texbook.
Start from calculation moles of NH₄Cl and NaOH.
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: RayFlacco on December 11, 2016, 05:25:09 PM: 0.0187 moles of NH4Cl and 0.01 of NaOH, is that right? Still don't know what the next step is
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: Arkcon on December 11, 2016, 05:33:37 PM: Quote from: RayFlacco on December 11, 2016, 04:11:05 PM
Yeah that's what I'm struggling with. Do you know somewhere I can read the steps or something?

You can read them in your textbook, or class notes, as AWK: suggested. That's part of the Forum Rules{click} (http://www.chemicalforums.com/index.php?topic=65859.0), and follow them. You agreed to read and follow them when you signed up, and you're not really doing that. We can't teach you all steps one posting at a time, and you are, in fact, asking for that. Its hard to convince people sometimes.

Now, you've determined moles you have, what are the moles specified in your blanced chemical reaction? That amount is there, its just not specified, but you're expected to know its there.
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: AWK on December 11, 2016, 06:12:30 PM: Now calculate reagents on the right side of reaction.
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: Enthalpy on December 13, 2016, 11:01:36 AM: Quote from: RayFlacco on December 11, 2016, 03:45:18 PM
:rarrow: NH₃(g) + H₂O(l)
When can this happen? With very little water?
And wouldn't NH₃ dissolve a bit more by letting NaCl precipitate?
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: AWK on December 13, 2016, 11:53:09 AM: From these numbers the concentration of NaCl is to low for precipitation (even taking into account a common ion effect). Of course, all ammonia will dissolve in solution forming a buffer solution with ionic strength almost 2 M.
Title: Re: Reaction between ammounium chloride and sodium hydroxide
Post by: Enthalpy on December 15, 2016, 02:20:41 PM: NH₄Cl is solid at the beginning, rather than gaseous.
With the given amounts, the reaction is mainly a dissolution of NH₄Cl in water.
Na⁺ plays no role and OH^- possibly a small one because ammonia is a weak base.
That is, the initial equation is irrelevant.
Would that be correct?