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Excess reactant

  1. Apr 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi, following Wednesday i'm going to produce ammonia in the lab at school but we have 3 questions we have to solve beforehand.
    The first one asks the global reaction being ammonium chloride reacting with calcium hydroxide to form ammonia, water and calcium chloride, but it also asks WHY this reaction happens. I don't really know how i could explain that.
    The next question asks me to determine the amount of reactant needed to get 6.40 L ammonia at 20 °C and 1 atm and we work with a 2x excess of Ca(OH)2.

    2. Relevant equations
    The equation is clearly 2 NH4Cl + Ca(OH)2 --> 2 NH3 + CaCl2 + 2 H2O. So we lose as much NH4Cl as we gain ammonia.
    3. The attempt at a solution
    From the ideal gas law i find i gain 0.266 moles of ammonia. Now comes the problem: can i assume NH4Cl is the limitant reactant here as i know Ca(OH)2 is the excess reactant? So i would say the starting amount of NH4Cl is 0.266 moles and as Ca(OH)2 is in 2x excess it's the double meaning 0.532 moles. Did i work correctly or did i completely miss something here? The main confusion to me is the excessive reactant and how to calculate with it. Thanks in advance!
    Last edited: Apr 23, 2017
  2. jcsd
  3. Apr 23, 2017 #2


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    Staff: Mentor

    It will be much easier to measure amount of ammonium chloride needed than amount of calcium hydroxide - the latter typically containing unknown amounts of water.

    As to WHY - which is a stronger base, OH- or NH3?
  4. Apr 23, 2017 #3


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    Also what is the physical state of both reactants and products - would you expect that to play a part?
  5. Apr 24, 2017 #4
    Thanks for your answer! But i still need to determine the mass calcium hydroxide needed, and i'm quite sure it's allowed to neglect the contained water. So as calcium hydroxide is in 2x excess, does that just mean i start with twice as much Ca(OH)2 as NH4Cl or is that wrong?
    Last edited: Apr 24, 2017
  6. Apr 24, 2017 #5


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    It is rather confusing what your reasoning is and what you are supposed to be working towards.
    I don't know quite how you calculated 0.266 moles ammonia, – I calculate about 0.29, but we're roughly in the same area.
    I don't know if you have noticed it but your own equation is saying that a mole of Ca(OH)2 gives you 2 moles of ammonia.
    So you do have a factor 2 in there, but it's the other way round from what you say and in fact I don't know whether that has anything to do with the question, nor where your 2× factor comes from. Were the problems given you in writing, and can you quote it all exactly?
  7. Apr 25, 2017 #6
    Given questions:
    1) Write out the equation of the happening reaction (written out in my first thread)
    2) Calculate the mass of ammonium chloride and calcium hydroxide needed to produce 6.40 L of ammonia at 20.0 °C and 1 atm. We work with an excess amount (2x) of Ca(OH)2.

    So with these given numbers i calculated the amount of ammonia gained in moles with the ideal gas law, that i found to be 0.266 moles. So now i would use that amount to calculate using stoichiometry the number of moles of reactants i had. And as i know from the given exercise, Ca(OH)2 is in an excess amount i assume to be the double amount of moles NH4Cl. So i also assume NH4Cl is the limitant reactant so i had 0.266 moles of ammonium chloride at the beginning meaning 14.2 grams. If i double it for calcium hydroxide it becomes 0.522 moles which means 39.5 grams, but that seems a lot to me. So my question is did i work correctly with that (2x), or does it mean something else to be taken into account ? Did i do right by assuming that (2x) excess means i start with double as much moles for calcium hydroxide as ammonium chloride ?
    Hope i cleared it up :-)
  8. Apr 25, 2017 #7


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    More than ideal gas laws I guess you're using the ideal gas constant R. I wonder what temperature you are assuming? I just remember that a mole of ideal gas is 22.4 L at "NTP", but perhaps the terminology is no longer much used. Anyway this is for now a fairly minor point.
    Assuming 0.266 moles ammonia is what you want then 14.2 g ammonium chloride is the right amount.

    You did not do the rest of the calculation right. Forget about the twofold excess required (it's not going to make much difference I think whether you put in a twofold excess or a tenfold excess). First calculate how much calcium hydroxide is exactly equivalent to that of the ammonium chloride. You are ignoring what I reminded you of in the last post about how many moles of calcium hydroxide gives you how many moles of ammonia, i.e. what your equation in #1 says.

    You are also ignoring the question and my hint about why this reaction goes. Do you know anything about chemical equilibria?
    Last edited: Apr 26, 2017
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