Respuesta :
Answer:
5.004kg
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: 4999.98 kg
Explanation:
Balanced reaction for combustion of carbon:
[tex]C+O_2\rightarrow CO_2[/tex]
Moles =[tex]\frac{\text{ given mass}}{\text{ molar mass}}[/tex]
Moles of [tex]CO_2=\frac{1.00\times 1000\times 1000g}{12g/mole}=83,333moles[/tex]
From the stoichiometry:
1 mole of C produces 1 mole of [tex]CO_2[/tex]
Thus 83333 moles of C produce=[tex]\frac{1}{1}\times 833333=83333moles[/tex] of [tex]CO_2[/tex]
From the combustion reaction equation provided
[tex]CO_2(g)+2NH_3(g)\rightarrow CO(NH_2)_2(s)+H_2O(l)[/tex]
From the stoichiometry:
1 mole of [tex]CO_2[/tex] produces = 1 mole of [tex]CO(NH_2)_2[/tex]
Thus 83,333 moles of [tex]CO_2[/tex] produce=[tex]\frac{1}{1}\times 83,3333=83,333moles[/tex] of [tex]CO(NH_2)_2[/tex]
Mass of [tex]CO(NH_2)_2=moles\times {\text {Molar mass}}=83333\times 60=4999980g=4999.98kg[/tex] As (1kg=1000g)
Thus maximum mass of urea that can be manufactured from the [tex]CO_2[/tex] produced by combustion of [tex]1.00\times 10^3kg[/tex] of carbon is 4999.98 kg
