Question

An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia with excess carbon dioxide. Determine the theoretical yield of urea and percent yield for the reaction.

Answer 1

The theoretical yield of urea is 93.05 kg, and the percent yield for the reaction is 94.04%.

Step-by-step explanation:

Theoretical Yield:

First, we need to balance the chemical equation:

CO₂(g) + 2NH₃(g) → CO (NH₂)₂(s) + H₂O (l)

From the balanced equation, we can see that for every 1 mole of CO₂, we obtain 1 mole of urea (CO(NH₂)₂).

Molar mass of CO₂ = 44.01 g/mol

So, mass of CO₂ in 68.2 kg = 68.2 kg × (1000 g/ 1 kg) × (1 mol/44.01 g) = 1549.41 mol

Thus, the maximum theoretical yield of urea = 1549.41 mol × (60.06 g/ 1 mol) = 93,046.5866 g = 93.05 kg

Percent Yield:

Percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100%:

Percent yield = (87.5 kg/ 93.05 kg) × 100% = 94.04%

Answer 2

The theoretical yield of urea = 120.35kg

The percent yield for the reaction = 72.70%

Step-by-step explanation:

Lets calculate -

The given reaction is -

`2NH_3(aq)+CO_2` →
`CH_4N_2O(aq)+H_2O (l)`

Molar mass of urea
`CH_4N_2O`= 60g/mole

Moles of
`NH_3` =
`(62.8kg/mole)/(17g/mole)` (since
`Moles=(mass of substance)/(mass of one mole)`)

= 4011.76 moles

Moles of
`CO_2` =
`(105kg)/(44g/mole)`

=
`(105000g)/(44g/mole)`

= 2386.36 moles

Theoritically , moles of
`NH_3` required = double the moles of
`CO_2`

but ,
`4011.76<2* 2386.36` , the limiting reagent is
`NH_3`

Theoritical moles of urea obtained =
`(1 mole CH_4N_2O)/(2mole NH_3)*4011.76 mole NH_3`

=
`2005.88mole CH_4N_2O`

Mass of 2005.88 mole of
`CH_4N_2O` =
`2005.88 mole *(60g CH_4N_2O)/(1mole CH_4N_2O)`

= 120352.8g

`120352.8g* (1kg)/(1000g)`

= 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild =
`(87.5kg)/(120.35kg)*100`

72.70%

Thus , the percent yeild for the reaction is 72.70%