How do I find the concentration of the excess reagent?
I already have the moles of excess reagent. I'm stuck because I don't know how to find the volume of excess (I'm not given a density).
concn = moles/L
If you have moles and you have the volume of the solution that's all you need.
My excess is HNO3. For the reaction, I added 15 mL of 6.00 mol/L HNO3. And then I found the moles of excess.
My concentration of excess is not 6.00 mol/L, is it? Don't I have to use the moles of excess that I found?
Sorry, but do you mean I take the moles of excess that I calculated and divide it by 15 mL?
You know it would be MUCH more simple if you typed in the whole question. As it is we are guessing.
Sorry! Here it is:
"Copper metal is reacted with concentrated nitric acid to form copper(II)nitrate in solution. Water and nitrogen dioxide gas are also produced in the reaction."
"What is the concentration of the excess reagent?"
As it is the problem can't be solved; You must have been given a mass of Cu and perhaps 15 mL of 6.00 M HNO3. Put all of the problem in please.
As mentioned above, 15 mL is added and the concentration of nitric acid is 6.00 moL/L.
Oh, also, I started off with 1.00 g of copper.
(Sorry I'm forgetting to add important information. It's a lab procedure, and so the information is all over the place.)
Cu + 4HNO3 ==> Cu(NO3)2 + 2NO2 + 2H2O
You will need to do it more exactly because I'm rounding here and there for the atomic masses and molar masses.
moles Cu = 1/63.5 = about 0.0157 moles.
moles HNO3 = M x L = 6.00 x 0.015 L = 0.090.
The reaction tells us that it will take 4 moles HNO3 for every mole of Cu; therefore, moles Cu x 4 = moles HNO3.
0.0157 x 4 = 0.063.
We had 0.090 initially, we used 0.063 so we have left 0.090-0.063 = 0.027 moles HNO3 excess.
Conn = moles/L = 0.027 moles/0.015 L = ??
Check my work. Check the equation to make sure it is balanced. About 1.8 M for HNO3 left?