Respuesta :
Answer:
43.13Kg of melamine
Explanation:
The problem gives you the mass of urea and two balanced equations:[tex]CO(NH_{2})_{2}_{(l)}=HNCO_{(l)}+NH_{3}_{(g)}[/tex]
[tex]6HNO_{l}=C_{3}N_{3}(NH_{2})_{3}_{(l)}+3CO_{2}_{(g)}[/tex]
First we need to calculate the number of moles of urea that are used in the reaction, so:
molar mass of urea = [tex]60.06\frac{g}{mol}*\frac{1kg}{1000g}=0.06006\frac{Kg}{mol}[/tex]
The problem says that you have 161.2Kg of urea, so you take that mass of urea and find the moles of urea:
161.2Kg of urea[tex]*\frac{1molofurea}{0.06006Kgofurea}=[/tex]2684 moles of urea
Now from the stoichiometry you have:
2684 moles of urea[tex]*\frac{1molofHNCO}{1molurea}*\frac{1molofmelamine}{6molesofHNCO}[/tex] = 447 moles of melamine
The molar mass of the melamine is [tex]126.12\frac{g}{mol}[/tex] so we have:
[tex]447molesofmelamine*\frac{126.12g}{1molofmelamine}[/tex] = 5637.64 g of melamine
Converting that mass of melamine to Kg:
5637.64 g of melamine *[tex]\frac{1Kg}{1000g}[/tex] = 56.38 Kg of melamine, that is the theoretical yield of melamine.
Finally we need to calculate the mass of melamine with a yield of 76.5%, so we have:
%yield = 100*(Actual yield of melamine / Theoretical yield of melamine)
Actual yield of melamine = [tex]\frac{76.5}{100}*56.38Kg[/tex] = 43.13Kg of melamine
