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Excess reactant

  1. Apr 25, 2006 #1
    ok...i know how to find the limiting reactant but am lost as in how to find out how much is left over...

    Example:

    N2H4 + 7H2O2 ---> 2HNO3 + 8H2O

    a) 120g of N2H2 reacts with an equal mass of H2O2. Which is the limiting reactant.

    i got is done:
    N2H2= M/mm=120g / 32.1 g/m = 3.74 moles
    H2O2= M/mm=120g / 34.02 g/m = 3.5 moles

    hence H2O2 is the limiting reactant.

    b) What mass of HN03 is expected?

    2/x = 7/3.5
    7x=7
    x=1 mole

    HN03= 63.02 g/mole X 1 mole = 63.0 g

    C) What mass, in grams, of the excess reactant remain at the end of the reaction?

    this is where im lost...some help would be nice...
     
  2. jcsd
  3. Apr 25, 2006 #2

    mrjeffy321

    User Avatar
    Science Advisor

    In part a), to find the limiting reactant you found the number of moles present of each of the reacts.

    N2H2= M/mm=120g / 32.1 g/m = 3.74 moles
    H2O2= M/mm=120g / 34.02 g/m = 3.5 moles

    You know from the chemical equation that the N2H4 reacts in a 1:7 ratio with H2O2,
    N2H4 + 7H2O2 --->
    1 mole of N2H4 : 7 moles H2O2

    If you know that H2O2 is the limiting reactant, you know that after the reaction, none remains. All the H2O2 needed to react with the excess reactant, N2H4 in its propper molar ratios. For every 3.5 moles of H2O2 reacted, .5 moles of N2H4 are needed (3.5 / 7). Thus, after the reaction, you have .5 moles less of N2H4 than you did when you started, for a remaining 3.24 moles.
    Converting this number of moles to grams is simply a matter of multiplying by the molar mass.
     
  4. Apr 25, 2006 #3
    thanks i got it...just a bit confused...now its clear...thanks again :wink:
     
  5. Apr 26, 2006 #4
    Can I get some assistance with concentration problems?

    1. What mass of formaldehyde, CH20 is contained in 250 mL of 8.0 mol/L solution of formaldehyde.

    I went about this with first getting the molar mass of formaldehyde; which is 30 g/mol. Then since 8.0 mole/L I divided it by 4 to get 2 mol/250 mL.

    Mas= MM x moles
    =30 g/mol x 2.0
    =60 g

    Is this correct?


    2. Calculate the concentration of sucrose solution when 5.0 g of sucrose, C^12 H^22 O^11, is added to water make 50.0 mL of solution.

    I did this in a mass/volume percent formula.

    Mass/volume= (mass of solute(g)/Volume of solution) X100%
    =(5.0g /50.0mL)X100%
    =10g/ml

    Correct?


    3. What volume of a 6.0 M solution of potassium nitrate would contain 3.0g of solute.

    I am stuck hear a bit. Do I need to use molar mass of potassium to get its mass and then use a mass/mass formula to get the answer?


    4. Stock solutions of sulphuric acid are 96.0 % H2SO4 by mass. These stock solutions have a density of 1.84 g/mL. Calculate the concentration of these stock solutions of sulphuric acid in mol/L.


    I need mad help to start these off...am a bit thrown off...dont know where to start...


    5. Stock solutions of hydrochloric acid have a concentration of 11.66 mol/L and a density of 1.18
    g/mL. Calculate the concentration in units of mass/mass percent of these stock solutions.

    same as 4...need a place to start...
     
    Last edited: Apr 26, 2006
  6. Apr 27, 2006 #5

    Borek

    User Avatar

    Staff: Mentor

    Last edited by a moderator: Aug 13, 2013
  7. May 8, 2006 #6

    For concentration problems u must know the almighty formulae:
    No. of moles= Concentration x Volume (when the volume is given in L or dm^3)
    No. of moles=Concentration x Volume/1000 (when the volume is in mL or cm^3)
    And also, No. of moles=Mass/Molar Mass
    Then ur all set!

    1. First of all ur solution is not correct.:surprised

    No. of moles= C x V/1000
    =8 x 250/1000
    =2 moles.

    Molar mass of CH20 = 12 + (1 x 20)= 32 g/mol

    No. of moles=Mass/Molar Mass
    Mass=No. of moles x Molar Mass
    =2 x 32=64g

    2. I am really surprised at the way u approach to solve these problems......I really wonder how u come up with the concepts u stated....read a couple of good books in chemistry..........

    Molar Mass of sucrose= (12 x 12) + (1 x 22) + (16 x 11)= 342g/mol

    No. of moles of sucrose=5/342=0.0146 mol

    No. of moles= C x V/1000
    0.0146= C x 50/1000
    C=0.292mol/L

    3. The molecular formula for potassium nitrate is KNO3.
    Molar mass of KNO3= 39 + 14 + (16 x 3) = 101g/mol

    No. of moles of KNO3=3/101=0.0297 mol

    No. of moles= C x V/1000
    0.0297 = 6 x V/1000
    V=4.95 mL

    Well i gotta go to sleep now:zzz: , the rest is for u to find out..........
    ***B***
     
  8. Dec 29, 2008 #7
    Is it possible for you to show this in a calculation formate, because em not getting the meaning behind the words :| PLEASE AND THANKS! <3
     
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