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Limiting Reactants

Posted: Wed Oct 03, 2018 5:20 pm
by Ariana Morales
Hi, I was wondering if anyone could help me with M.9 (7th edition). Specially, I need help setting up the final balanced equation.
The question states: "Copper(II) nitrate reacts with sodium hydroxide to produce precipitate of light blue copper(II) hydroxide."
a) Write the net iconic equation for the reaction?

Any clarification will be highly appreciated.

Re: Limiting Reactants

Posted: Wed Oct 03, 2018 5:34 pm
by armintaheri
First thing to notice is that it asks for the "net ionic equation." That means you only include reactants and products that are involved in a permanent or irreversible reaction. If you remember solubility rules, you know that any ionic compound containing nitrate or sodium is soluble in water. So a nitrate ion and a sodium ion might combine to form sodium nitrate. But since sodium nitrate is very soluble, it's going to immediately break apart again. So we don't include sodium and nitrate in our equation since they don't participate in an irreversible reaction. They don't make a permanent product so we don't care about them. The only ions we care about are copper and hydroxide. You know copper(II) has a charge of 2+ from the roman numerals in the parentheses. You know hydroxide always has a charge of -1. You also know that any ionic compound must be neutral. So to get neutral copper hydroxide, we would need one copper ion for every two hydroxide ions. So the final equation is

Re: Limiting Reactants

Posted: Wed Oct 03, 2018 5:40 pm
by arif_latif_2G
First, you should figure out the molecular formulas of the products and reactants.
Reactants: Cu(NO3)2 and NaOH
Products: Cu(OH)2 and NaNO3 (not mentioned within the question)
So you get Cu(NO3)2 + NaOH -> Cu(OH)2 + NaNO3
Now to balance the equation: On the reactant side, you can see that you have 2 nitrate ions, so in order to balance it, you put a 2 in front of NaNO3 so you have 2 nitrate ions on the product side. From there you will notice that you now have 2 Na atoms on the product side, you put a 2 in front of NaOH to balance on the reactant side and from there you will see that the Copper and Hydroxide ions will be balanced as well.

So you end up with: Cu(NO3)2 + 2NaOH -> Cu(OH)2 + 2NaNO3

If you want to know the net ionic equation, separate the ions in the equation: (Cu2+) + (2NO3-) + (2Na+) + (2OH-) -> (Cu(OH)2) + (2Na+) + (2NO3-) **note that Cu(OH)2 remains the same because it is a precipitate solid and cannot be separated into ions in its current state.

Then you can cancel the ions that show up on both sides of the product and reactant sides of the reaction: (2Na+) + (2NO3-), leaving you with (Cu2+) + (2OH-) -> (Cu(OH)2)

Re: Limiting Reactants

Posted: Wed Oct 03, 2018 5:58 pm
by Ariana Morales
Thank you, it was very helpful!