posted by .

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).

  • Chemistry -

    concn = moles/L
    If you have moles and you have the volume of the solution that's all you need.

  • Chemistry -

    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?

  • Chemistry -

    Sorry, but do you mean I take the moles of excess that I calculated and divide it by 15 mL?

  • Chemistry -

    You know it would be MUCH more simple if you typed in the whole question. As it is we are guessing.

  • Chemistry -

    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?"

  • Chemistry -

    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.

  • Chemistry -

    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.)

  • Chemistry -

    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?

  • Chemistry -

    Thank you so much for taking the time to work through it all...and for your patience! I understand now!

  • Chemistry -

    You're welcome.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    If 12.50 grams of Pb(NO3)2 are reacted with 3.18 grams Nacl in a "typical double displacement" reaction, please calculate the following: a) What is the limiting reagent?
  2. orgenic chemistry

    which starting material is the limiting reagent in this procedure?
  3. Chemistry

    Thank you Dr Russ for you answer. Anyone who can answer: I have the moles of excess reagent, but now I need to find the concentration of excess reagent. I know concentration is mol/L, but how do I find the volume of my excess?
  4. Chemistry!

    10.2 g of Pb(NO3)2 are dissolved in 100 mL of water, and added to 250.0 mL of a 0.250 M solution of KI. Determine the mass in grams of Pbl2 produced, and find the number of moles of the excess reagent. PLZ explain chemistry - DrBob222, …
  5. chemistry

    17.8 grams of CH4 reacts with 55.4 g of )2 to from CO2 and H20 what is the mass of H20 formed?
  6. Chemistry

    320g of sulfur dioxide react with 32.0g of oxygen and excess water to form sulfuric acid (H2SO4) a. What is the limiting reagent?
  7. Chemistry

    Based on the balanced equation 2NBr3 + 3NaOH → N2 + 3NaBr + 3HOBr calculate the number of of excess reagent units remaining when 116 NBr3 molecules and 180 NaOH formula units react?
  8. Chemistry

    Al_S=Al2S3 10 grams of Al were mixed with 15 gram of S Find: limiting reagent al or s?
  9. chemistry

    Greg reacts 5.0 moles of hydrochloric acid(HCL)with 1.5 moles of barium hydroxide(Ba(OH)2).Which reagent is the limiting reagent, and which is in excess?
  10. Chemistry

    7.0 mL of 0.64 M HBr reacts with 38 mL of 0.39 M KOH. 1) What are the moles of acid before the reaction?

More Similar Questions