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Calculating the excess reagent remaining at the end of the reactionhelp?

  1. Dec 6, 2009 #1
    Calculating the excess reagent remaining at the end of the reaction..help!?!

    SO here's the question:

    Q: Copper (II) Sulfide (0.600mol) is treated with 1.40mol of nitric acid.

    Here's the equation: 3 CuS + 8 HNO3 --> 3 Cu(NO3)2 + 3 S + 2 NO + 4 H2

    A) How many moles of Copper (II) Nitrate can be produced?
    `i already answered this part :D
    For CuS, .600 moles of Copper II Nitrate will be produced, and for Nitric acid...0.525 moles of Copper II Nirate will be produced.

    B) If 0.500 mol of Copper II Nitrate is actually obtained, what is the percent yield?
    `for this, I got 95.2% :D

    C) Calculate the # of moles of excess reagent remaining at the end of the reaction.
    `help me with this please. how do i calculate the excess reagent. The excess reagent is CuS, ryyyt =]
     
  2. jcsd
  3. Dec 6, 2009 #2

    Lok

    User Avatar

    Re: Calculating the excess reagent remaining at the end of the reaction..help!?!

    A) Which one is it? 0.6 or 0.525 they cannot be both right.

    B) you used the 0.525 so you got it

    C) How much does remain after you take 0.525 theoretical (or 0.5 practical value)out of what you had at first?
     
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