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Offline NullSector

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Theoretical/Actual Yield
« on: February 01, 2024, 01:22:54 PM »
Hey all. I just needed some help verifying that this is correct. Thanks all!

   Use the following equation to answer the questions below:
CH4 + NH3 + O2 > HCN + H2O
   Balance the equation.
   Given that we have 21 g of methane (CH4) and 12 g of ammonia (NH3), which reactant is limiting? Show your calculations and explain how you determined the limiting reactant.
   What is the theoretical yield of hydrogen cyanide will be formed? Show all your work. 
   If the actual yield of hydrogen cyanide is 53%, then how many grams of HCN is produced?
   What is the actual yield of water?

Answer:
   The balanced equation will be 2CH_4+2NH_3+3O_2→2HCN+6H_2 O
   
Given that the ratio of reactants to product in the balanced equation is 2:2:3→2:6, the limiting reactant would be ammonia, NH_3. The ratio of 2:2:3→2:6 states that 2 moles of CH_4, 2 moles of NH_3, and 3 moles of O_2 will produce 2 moles of HCN and 6 moles of H_2 O.
Find how many moles of each reactant we have by determining the molar weight.
CH_4  moles=21/[C+(H*4)]
CH_4  moles=21/[12+(1*4)]
CH_4  moles≈1.3

NH_3  moles=12/([N+(H*3)])
NH_3  moles=12/([14+(1*3))
NH_3  moles≈0.7

The moles of ammonia and methane should be equal for there to be no limiting reactant, but as there is less moles of ammonia, it is the limiting reactant.
   

To find the theoretical yield of HCN, first find how many moles of the limiting reactant is present in 12g. Since the limiting reactant is NH_3, find its molar weight, in g/mol.
The weight of N is about 14 amu. The weight of H is about 1 amu. Combine: 14+(1*3)=17 g/mol.
Divide 12 by 17: 12/17≈0.7
Since the ratio of NH_3 to HCN is 2:2, or 1:1, 0.7 moles of NH_3 will produce 0.7 moles of HCN.
Find the molar weight of HCN. The mass of hydrogen (H) is about 1 amu. The mass of carbon (C) is about 12 amu. The mass of nitrogen (N) is about 14 amu. Combine: 1+12+14=27g/mol.
Multiply the molar mass by the number of moles: 0.7*27=18.9g.
Thus, there is about 18.9 grams produced with a theoretical yield of 100%.
   

If the actual yield of hydrogen cyanide is 53%, when provided with 2 moles of NH_3, 2*0.53 or 1.06 moles of HCN will be produced. The ratio is now 2:1.06
As we are given 0.7 moles of NH_3, multiply it by the fraction 1.06/2  to find the moles produced of HCN: 0.7*1.06/2=0.371. Thus, 0.371 moles of HCN are produced with a yield of 53%.
Multiply the moles by the molar weight of HCN: 0.371*27=10.017.
We could also find the actual yield by multiplying the theoretical yield (18.9) by 53% (0.53): 18.9*0.53=10.017.
Thus, there are about 10.017 grams of hydrogen cyanide produced with a yield of 53%
   

The ratio of the first product, HCN, to H_2 O is 2:6,  or 1:3.
To find the actual yield of water when given the actual yield of hydrogen cyanide (10.017 grams), multiply the amount of HCN by 3.
10.017*3=30.051
Thus, there is about 30.051 grams of actual water yielded.  

Offline Hunter2

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Re: Theoretical/Actual Yield
« Reply #1 on: February 01, 2024, 01:50:54 PM »
Looks everything ok.

Offline Borek

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Re: Theoretical/Actual Yield
« Reply #2 on: February 01, 2024, 02:55:33 PM »
NH_3  moles≈0.7

As you wrote: this is an approximate value, and 0.7 is a rounded down number of moles. Never use rounded down values for calculations. Report them rounded down, but when you continue calculations use full precision (or at least use several so called guard digits).

Exact expected mass of HCN for 100% yield is 19.0 g, not 18.9 g. Small difference, but 1. you are losing accuracy, 2. if you do the same (round down) several times in a row for several intermediate steps/values, these errors can accumulate and you can get off by quite substantial amount.
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Offline NullSector

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Re: Theoretical/Actual Yield
« Reply #3 on: February 01, 2024, 05:12:13 PM »
Thanks all!

Offline NullSector

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Re: Theoretical/Actual Yield
« Reply #4 on: February 01, 2024, 05:13:29 PM »
Thank you...my teacher wanted it rounded down from the atomic mass for some reason, but ill keep that in mind for future courses
NH_3  moles≈0.7

As you wrote: this is an approximate value, and 0.7 is a rounded down number of moles. Never use rounded down values for calculations. Report them rounded down, but when you continue calculations use full precision (or at least use several so called guard digits).

Exact expected mass of HCN for 100% yield is 19.0 g, not 18.9 g. Small difference, but 1. you are losing accuracy, 2. if you do the same (round down) several times in a row for several intermediate steps/values, these errors can accumulate and you can get off by quite substantial amount.

Offline mjc123

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Re: Theoretical/Actual Yield
« Reply #5 on: February 01, 2024, 05:14:11 PM »
The last step is wrong. Themolar ratio of water to HCN is 3;1, yet you multiply the mass of HCN by 3.

Offline NullSector

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Re: Theoretical/Actual Yield
« Reply #6 on: February 01, 2024, 05:26:25 PM »
Oh my that you are correct I messed up when I did it.
Is this correct?:
   The ratio of the first product, HCN, to H_2 O is 2:6,  or 1:3.
To find the actual yield of water when given the actual yield of hydrogen cyanide (10.017 grams), multiply the moles of HCN by 3.
0.371*3=1.113
Thus, there is about 1.113 moles of actual water yielded.
Multiply by the molar mass to find the amount yielded.
mass=[(H*2)+O]
mass≈(1*2)+16
mass≈18 g/mol
18 g/mol * 1.113 moles
20.034 grams
There is about 20.034 grams of water produced.

Offline Borek

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Re: Theoretical/Actual Yield
« Reply #7 on: February 01, 2024, 06:21:47 PM »
Thank you...my teacher wanted it rounded down from the atomic mass for some reason, but ill keep that in mind for future courses

To reiterate: it is OK and correct to report rounded down intermediate values (although in this case it would be better to report as 0.70, using two significant figures in accordance with the data given), just don't use rounded values in the following calculations.

There is about 20.034 grams of water produced.

Yes (although again, significant figures suggest it would be better to report it as 20 g).
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