Stoichiometry applied to xN2 + yH2

  • Thread starter rijo664
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  • #1
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6.0 moles of N2 thats are mixed with 12.0 moles of H2 according to the following equation:

The equation is already balanced:

N2(g)+ 3H2(g) yields 2NH3 (g)

The questions asks for what is the chemical in excess? is it H2 or N2
I solved the problem but i want to make sure if it is right. this is what i did

6.0 moles of N2 (2 mol NH3 / 1 Mol of N2)= 12 mol of NH3
12.0 moles of H2 (2 mol NH3 / 3 mol H2)= 8 mol of NH3

I believe the answer is N2 because given amount is 12.0 moles and i got 8 mol of NH3 is it right or wrong please tell me.
 

Answers and Replies

  • #2
1,753
1
that is correct :-]]]
 
  • #3
25
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Let me ask u another question

Theoretically, how many moles of NH3 will be produced

I believe it's 2 Moles of NH3 because that's how many there are in the balanced equation. Is it right if so

How should i get the actual yield of NH3 if the percent yield is 80%, how many moles of NH3 are actually produced. I am kind of stuck here if the theoretical is 2 then how will i find the actual

I know the formula for percent yield is

actual yield/theoretical yield X 100

help would be appreciated.
 
  • #4
1,753
1
the Theoretical Yield is calculated by determining which is the limiting reagent and how much of the product it will produce

the formula that relates the Theoretical Yield and Actual Yield is, % Yield = Actual/Theoretical
 
Last edited:
  • #5
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I believe the reagent is N2 (that's N and the subscript 2)
 
  • #6
1,753
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I believe the reagent is N2 (that's N and the subscript 2)
sorry i meant the Limiting reagent
 
  • #7
25
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still i don't get it, if the theoretical yeild of NH3 is 2 mol and the limiting reagent is H2 how is that suppose to work i don't get it.
 
  • #8
1,753
1
[tex]N_2 (g) + 3H_2 (g) \rightarrow 2NH_3[/tex]

[tex]6.0 mol N_2 = \frac{2 mol NH_3}{1.0 mol N_2} = 12.0 mol NH_3[/tex]

[tex]12.0 mol H_2 = \frac{2 mol NH_3}{3.0 mol H_2} = 8.0 mol NH_3[/tex]

So, the since, Hydrogen gas produces the least amount of Ammonia, it is the Limiting Reagent. The Limiting Reagent also tells you the Theoretical Yield. The Theoretical Yield is simply the amount of Product that will be formed from the Limiting Reagent.

Thus, our equation that relates the 2 is:

[tex]\% Yield = \frac{Actual-Yield}{Theoretical-Yield}\times 100[/tex]

So we want to solve for Actual-Yield, it's basically plug and chug.
 
Last edited:

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