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Topic: Limiting Reactants and Percent Yield  (Read 11378 times)

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liffey07

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Limiting Reactants and Percent Yield
« on: August 13, 2005, 11:48:08 PM »
--Statement--
Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane:
2NH3(g) + 3O2(g) + 2CH4(g) --> 2HCN(g) + 6H2O(g)

--Question--
If 5.00 x 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?

--My process--
I assumed that one of the three reactants had to be limiting.  I first found the number of moles of each, which came out to be 293530.586 mol NH3, 156250 mol O2, and 311681.835 mol CH4.  Then I set up two ratios of available moles:

293530.586 mol NH3  = 1.8785957 mol NH3
      156250 mol O2             1 mol O2

311681.835 mol CH4  =  1.9947637 mol CH4
      156250 mol O2              1 mol O2

I compared these ratios to the ratios of required moles (derived from the balanced equation):

.6666667 mol NH3   .6666667 mol CH4
      1 mol O2                 1 mol O2

Because the available moles of NH3 and of CH4 were both in excess, I assumed that the O2 would be consumed first, and would therefore be the limiting reactant.  Next, I used the moles of the limiting reactant (O2) to find the masses of the products:

156250 mol O2x 2 mol HCN   x 27.028 g HCN = 2.82 x 103 kg HCN
                       3 mol O2        1 mol HCN

156250 mol O2  x 6 mol H2O  x 18.015 g H2O  = 5.63 x 103 kg H2O
                           3 mol O2          1 mol H2O

I thought I solved the problem correctly, but my solution cannot be valid because it does not follow the Law of Conservation of Mass; the mass of the reactants (5.00 x 103 kg * 3 = 1.5 x 104 kg) does not equal the mass of the products (8.45 x 103 kg).  I don’t know whether my approach to the problem is completely off.  For example, maybe I should have set up a third mol ratio comparing NH3 and CH4.  Also, I don’t know how the 100% yield will influence the problem.  It means that the theoretical yield and actual yield are equal, but does that mean all the reactants will be consumed (i.e., there will be no limiting reactant?).  I am just very confused and need some help. Thanks for your patience!  ;D

Offline xiankai

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Re:Limiting Reactants and Percent Yield
« Reply #1 on: August 14, 2005, 04:00:26 AM »
is that 5 x 103 or just 5 x 103?
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Offline Yggdrasil

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Re:Limiting Reactants and Percent Yield
« Reply #2 on: August 14, 2005, 04:40:52 AM »
Your answer is correct.  The reason the reaction seems to violate the conservation of mass is because the excess methane and ammonia cannot react to form hydrogen cyanide.  The missing 6.45x103 kg on the products is in the form of unreacted methane and ammonia.

Offline Donaldson Tan

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Re:Limiting Reactants and Percent Yield
« Reply #3 on: August 14, 2005, 04:05:31 PM »
... but my solution cannot be valid because it does not follow the Law of Conservation of Mass; the mass of the reactants (5.00 x 103 kg * 3 = 1.5 x 104 kg) does not equal the mass of the products (8.45 x 103 kg).  I don’t know whether my approach to the problem is completely off.  

the law of conservation of mass states that the mass of product formed equals the masss of reactants consumed.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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