chem please help?
Author: Tyler Evans
Question by kevin n: chem please help?
An experiment that led to the formation of the new field of organic chemistry involved the synthesis of urea, CN2H4O, by the controlled reaction of ammonia and carbon dioxide.
2 NH3(g) + CO2(g) CN2H4O(s) + H2O(l)
What is the theoretical yield of urea when 100. g of ammonia is reacted with 100. g of carbon dioxide?
Best answer:
Answer by mia_in_nz
First step is to find the relative molecular mass (Mr) for each molecule by adding the atomic masses:
Mr(NH3)= 14.01+ 3(1.008)
=17.034
At this point so i don’t confuse myself later, I times this by 2 since you need two moles to react with 1 mole of CO2
ie 36.068
Mr(CO2)= 12.01 + 2(16.00)
=46.01
Since you’re using the same amount of each compund it’s easy to see at this point that the CO2 will be the limiting reagent, since it’s Mr is greate, therefore the number of moles will be less.
The number of moles (n)= m/Mr
So:
n(NH3)= 100/36.068
=2.77moles
and
n(CO2)= 100/46.01
=2.17
Now the maximum amount of urea able to be produced by this reaction is the smallest number of moles times the Mr of urea
i.e.
2.17x [12.01+2(14.01)+4(1.008)+16]
=130.3g
Know better? Leave your own answer in the comments!
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Lancenigo di Villorba (TV), Italy
Wow! You shown a beautiful presentation of the Urea’s Synthesis!
HYSTORICAL NOTEs
Urea was discovered in second half of XVIII century from a frech medician, H. M. Rouelle.
At the middle of XIX century, chemists were able to detect it.
Urea was involved in increasing chemical experiments ; nonetheless, chemists need for dog’s kidneys since they ignored a different source.
F. Wohler was the german chemist which obtained urea from Inorganic Matters, e.g. Ammonium Cyanate or NH4OCN.
Nowadays, chemical industry follows the Ammonia-Carbon Dioxide Reaction in Closed-NotFlow’s Reactors.
CALCULATIONs
The balanced formula is the following one
2 NH3(l) + CO2(amm.) —> CO(NH2)2(amm.) + H2O(amm.)
since the Reaction’s Phase is a liquid one, e.g. Liquid Ammonia itself.
The stoichiometric correspondances follow
NH3 : CO2 : CO(NH2)2 = (-2) : (-1) : (+1)
where it appears that Ammonia is disappearing in a twice fashion than Carbon Dioxide does. This reasoning become expressed as follows :
nNH3 – nNH3° = 2 * (nCO2 – nCO2°)
nNH3 – (100 / 17) = 2 * (nCO2 – (100 / 44))
nNH3 – 5.9 = 2 * (nCO2 – 2.2)
nCO2 = 2.2 + 0.5 * (nNH3 – 5.9) mol
Starting from you cannot retrieve nCO2 < 0, the Limit Condition is the Carbon Dioxide’s Disappearance.
Thus, Carbon Dioxide’s Molar Amount will be zero and its Initial Value will be equal than Urea’s Molar Amount
nCO(NH2)2 = nCO2° = 2.2 mol
Urea’s Mass = Urea’s Moles * M.W.
Urea’s Mass = 2.2 * 60 = 132 g
I hope this helps you.
this is:
100*60/34=A
100*18/34=B
100*44/34=C
A=theoretical CN2H4O(g)
B=theoretical water
C=theoretical CO2 100(g)-C=not reaction (g)