Fundamental M.9

[Norah Gidanian 3D]

Copper(II) nitrate reacts with sodium hydroxide to produce a precipitate of light blue copper(II) hydroxide. (a) Write the net ionic equation for the reaction. (b) Calculate the maximum mass of copper(II) hydroxide that can be formed when 2.00 g of sodium hydroxide is added to 80.0 mL of
0.500 M Cu(NO3)2(aq)

Can someone walk me through how you would do this?

[John Pham 3L]

First, you need to write out the balanced chemical equation for the reaction.
- Cu(NO3)2 + 2NaOH --> Cu(OH)2 + 2NaNO3

For a net ionic equation, you need to only include the molecules that take part in the reaction.
To do this, you need to separate all molecules into ions and remove spectator ions.
- It is important to note that only Aqueous molecules can separate into ions

The Ionic Equation would then be Cu+2 + 2NO3- + 2Na+ + OH- --> Cu(OH)2 (s) + 2Na+ 2NO3-
Cu(OH)2 cannot dissociate in water so it would remain as a solid

Cancel out all spectator ions which are ions that are on both the reactant and product side of the reaction
You should be left with:
Cu+2 + 2OH- --> Cu(OH)2

Once you have the net ionic equation, you can then solve for part b
To calculate the maximum mass of copper (II) hydroxide, you need to convert all reactants into moles and see how much copper (II) hydroxide each could produce.
This will calculate the limiting reactant of the reaction and show the maximum mass of copper(II) hydroxide that could be produced

which could produce 2.44 g Cu(OH2) if there is sufficient Copper to react with

For Cu(NO3)2, you'll need to convert from molarity to moles
80.0 mL is equal to 0.0800 L
Multiply this by 0.500 M Cu(NO3)2 to get 0.0400 mol of Cu(NO3)2
Convert to Cu(OH2) and you would get 3.90 g Cu(OH)2 produced if there is sufficient Hydroxide to react with

Since 2.44 g < 3.90 g, then NaOH would be the limiting reactant
The maximum amount of Cu(OH)2 that could be produced would then be 2.44 g

[Grace_Remphrey_2J]

First, you need to write the net ionic equation. In a net ionic equation, you do not need to include the spectator ions. So, in this case, you do not need to include NO3 and Na. That means, in the end, the net ionic equation, will be Cu2+ + 2(OH)- yields Cu(OH)2.

In part b), you need to find the grams of Cu(OH)2 that can be formed. Because you are given the molarity of Cu(NO3)2, you will need to use the molarity formula (c=m/v) to find the number of mols of Cu(NO3)2 that are present (C stands for Molarity and V stands for volume in liters). After plugging everything in, you will find that there are .04 mols of Cu(NO3)2 and .5 mols of NaOH present. This is where you should determine the limiting reactant! NaOH is the limiting reactant because there is a 2 to 1 ratio between Cu2+ and 2(OH)-. You use the 2:1 ratio between 2(OH)- and Cu(OH)2 to find out that .025 mols of Cu(OH)2 will be formed. You use the molar mass of Cu(OH)2 to find the grams of Cu(OH)2 that are formed. In this case, that will be 2.44 g of Cu(OH)2.

[Leyla Anwar 3B]

Will ion equations be covered in Midterm 1? I realize it is part of the fundamentals so does that mean we should know how to solve these types of problems?