Question

Urea, CO(NH2)2, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO2 produced by combustion of 1.00 × 103 kg of carbon followed by the reaction? CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)

Answer 1

5.004kg

Step-by-step explanation:

Combustion of carbon

C+O2=CO2

from the relationship of molar ratio

mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)

mass of carbon =1000kg

atomic mass of carbon =12

volume of CO2 produced=1000×22.4/12

volume of CO2 produced =1866.6dm3

from the combustion reaction equation provided

CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)

applying the same relationship of molar ratio

no of mole of CO2=no of mole of urea

therefore

vol of CO2\22.4=mass of urea/molar mass of urea

molar mass of urea=60.06g/mol

from the first calculation

vol of CO2=1866.6dm3

mass of urea=1866.6×60.06/22.4

mass of urea=5004.82kg

Answer 2

Answer: 4999.98 kg

Step-by-step explanation:

Balanced reaction for combustion of carbon:

`C+O_2\rightarrow CO_2`

Moles =
`\frac{\text{ given mass}}{\text{ molar mass}}`

Moles of
`CO_2=(1.00* 1000* 1000g)/(12g/mole)=83,333moles`

From the stoichiometry:

1 mole of C produces 1 mole of
`CO_2`

Thus 83333 moles of C produce=
`(1)/(1)* 833333=83333moles` of
`CO_2`

From the combustion reaction equation provided

`CO_2(g)+2NH_3(g)\rightarrow CO(NH_2)_2(s)+H_2O(l)`

From the stoichiometry:

1 mole of
`CO_2` produces = 1 mole of
`CO(NH_2)_2`

Thus 83,333 moles of
`CO_2` produce=
`(1)/(1)* 83,3333=83,333moles` of
`CO(NH_2)_2`

Mass of
`CO(NH_2)_2=moles* {\text {Molar mass}}=83333* 60=4999980g=4999.98kg` As (1kg=1000g)

Thus maximum mass of urea that can be manufactured from the
`CO_2` produced by combustion of
`1.00* 10^3kg` of carbon is 4999.98 kg