Help with M9?

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janeane Kim4G
Posts: 31
Joined: Fri Sep 28, 2018 12:28 am

Help with M9?

Postby janeane Kim4G » Tue Oct 09, 2018 2:17 am

“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 max mass of copper (II) hydroxide that can be formed when 2.00g of sodium hydroxide is added to 80.0 mL of .500M Cu(NO3)2?”
How would you start off part B?

Gary Qiao 1D
Posts: 66
Joined: Fri Sep 28, 2018 12:26 am

Re: Help with M9?

Postby Gary Qiao 1D » Tue Oct 09, 2018 2:43 am

So first, you have the balanced equation: Cu(NO3)2 + 2NaOH --> Cu(OH)2 + 2NaNO3
The next step after this is to find the moles of the reactants in order to find the limiting reactant.
So for Cu(NO3)2:
n/L =M
n/.08L = .500M
n = .04 moles Cu(NO3)2

And for NaOH:
molar mass: 22.99g/mol Na + 16 g/mol O + 1.008g/mol H = 39.998 g/mol NaOH
2.00g/39.998g/mol = .05 moles NaOH

So .04 moles of Cu(NO3)2 would react with .05/2 = .025 moles of NaOH, making NaOH the limiting reactant.
With the balanced equation, using the ratios, you can find that the .05 moles of NaOH used up in the equation would produce .025 moles of Cu(OH)2 (since the ratio for NaOH to Cu(OH)2 is 2 to 1).

.025 moles Cu(OH)2 x 97.57g/mol Cu(OH)2 = 2.44 g Cu(OH)2

Hope this helps!

janeane Kim4G
Posts: 31
Joined: Fri Sep 28, 2018 12:28 am

Re: Help with M9?

Postby janeane Kim4G » Tue Oct 09, 2018 9:22 am

Thank you! This really helped


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