Hi Canyonhikers,
We should get the amount of ammonia and co2 reacting for the successful calcution of
Urea.
Suppose we have 100g of ammonia and 100g of CO2, the solution will be as follows.
.
We have 100g of ammonia, which is 5.88 moles of ammonia (100g x mole/17g).
We have 100g of CO2, which is 2.27 moles of CO2 (100g x mole/44g)
In order to consume 2.27 moles of CO2, we need 4.55 moles of ammonia (2 x 2.27, as
per above discussion). Once 4.55 moles of ammonia are consumed, any remaining [in
this case, since the problem states we have 5.88 moles (100g) available] ammonia just
sits around un reacted.
So we are limited by the amount of CO2 available, and for each mole of CO2 consumed,
we generate one mole of urea. So we expect to generate, theoretically, 2.27 moles of urea
here.
2.27 moles of urea (MW = 60g/mole) is therefore136g of urea. So the theoretical yield of
urea in this case is 136g.
Therefore 100gm ammonia gives 136 gm urea if the yield is 100%.
So for 5 gm Urea you need 3.676 gm of ammonia if 100g of ammonia and 100g of
CO2 in 100% yield
Option 2
If we are using
200 [g NH(3)] ) / (17 [g/mol] ) = 11.74 mol NH(3)
100 [g CO(2)] ) / (44 [g/mol] ) = 2.27 mol CO(2)
The equation tells you there are
2 moles NH(3) to 1 mole CO(2) = 2 [mol NH(3)] / [mol CO(2)]
From the actual mass given, there is actually
11.74 moles NH(3) to 2.27 moles CO(2) =
11.74 / 2.27 = 5.16 [mol NH(3)] / [mol CO(2)]
Since 5.16 > 2, there is an excess of NH(3), and CO(2) is the limiting reactant.
Therefore, use the 2.27 mol CO(2) to find the theoretical yield of urea:
2.27 [mol CO(2)] * 1 / 1 [mol CN(2)H(4)O] / [mol CO(2)]
= 2.27 mol CN(2)H(4)O
Multiply the # of moles of urea by the molar mass:
2.27 [mol CN(2)H(4)O] * 60 [g/mol] =136g urea
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