Respuesta :
Answer:
The theoretical yield of urea = 120.35kg
The percent yield for the reaction = 72.70%
Explanation:
Lets calculate -
The given reaction is -
[tex]2NH_3(aq)+CO_2[/tex] →[tex]CH_4N_2O(aq)+H_2O (l)[/tex]
Molar mass of urea [tex]CH_4N_2O[/tex]= 60g/mole
Moles of [tex]NH_3[/tex] = [tex]\frac{62.8kg/mole}{17g/mole}[/tex] (since [tex]Moles=\frac{mass of substance}{mass of one mole}[/tex])
= 4011.76 moles
Moles of [tex]CO_2[/tex] = [tex]\frac{105kg}{44g/mole}[/tex]
= [tex]\frac{105000g}{44g/mole}[/tex]
= 2386.36 moles
Theoritically , moles of [tex]NH_3[/tex] required = double the moles of [tex]CO_2[/tex]
but , [tex]4011.76<2\times 2386.36[/tex] , the limiting reagent is [tex]NH_3[/tex]
Theoritical moles of urea obtained = [tex]\frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3[/tex]
= [tex]2005.88mole CH_4N_2O[/tex]
Mass of 2005.88 mole of [tex]CH_4N_2O[/tex] =[tex]2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}[/tex]
= 120352.8g
[tex]120352.8g\times \frac{1kg}{1000g}[/tex]
= 120.35kg
Therefore , theroritical yeild of urea = 120.35kg
Now , Percent yeild = [tex]\frac{87.5kg}{120.35kg}\times100[/tex]
72.70%
Thus , the percent yeild for the reaction is 72.70%
