Need help on question M.9 for chem 14a

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OliverPark1C
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Joined: Fri Sep 25, 2015 3:00 am

Need help on question M.9 for chem 14a

Postby OliverPark1C » Wed Sep 30, 2015 5:45 pm

Hey I need help figuring out the question:
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).
Thanks!

Allison Yoon 3B
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Joined: Fri Sep 25, 2015 3:00 am

Re: Need help on question M.9 for chem 14a

Postby Allison Yoon 3B » Wed Sep 30, 2015 7:27 pm

First of all, 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. Cu2+ and 2(OH)- are both aqueous while Cu(OH)2 are solid. 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 c=m/v to find the number of mols of Cu(NO3)2 that are present. After plugging everything in, you will find that there are .04 mols of Cu(NO3)2. (C stands for Molarity and V stands for volume in liters.) You are also given the grams of NaOH. Use the molar mass of NaOH to find the number of mols of NaOH. There are .05 mols of NaOH present. It becomes evident that 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.

Kendall 3H
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Joined: Wed Sep 18, 2019 12:18 am

Re: Need help on question M.9 for chem 14a

Postby Kendall 3H » Mon Oct 07, 2019 10:04 pm

Thank you so much! I was very confused on this question and you explained it so well. I now see that you must first discover which reactant is the limiting reactant and then the problem becomes much more simple.

LBacker_2E
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Re: Need help on question M.9 for chem 14a

Postby LBacker_2E » Wed Oct 09, 2019 4:30 pm

In general, when you are writing a net ionic equation (like in this problem), how do you know which ions are spectator ions? Since some of the ions are going to be spectator ions, do those ions still need to be balanced on both sides of the equation?

Jack Riley 4f
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Joined: Sat Aug 24, 2019 12:17 am

Re: Need help on question M.9 for chem 14a

Postby Jack Riley 4f » Wed Oct 09, 2019 4:34 pm

To determine if an ion is a spectator ion you just need to look at the product that is stated in the problem. In this problem the product is copper(ii) hydroxide so you know that the sodium and nitrate are the spectators. Because they are spectators, they don't need to be balanced because they exist in the same form on both sides of the equation.


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