Question

Melamine, C3N3(NH2)3, is used in adhesives and resins. It is manufactured in a two-step process: CO(NH2)2(l) → HNCO(l) + NH3(g) 6 HNCO(l) → C3N3(NH2)3(l) + 3 CO2(g) What mass of melamine, C3N3(NH2)3, will be obtained from 161.2 kg of urea, CO(NH2)2, if the yield of the overall reaction is 76.5 % ?

Answer 1

Answer:

43.13Kg of melamine

Explanation:

The problem gives you the mass of urea and two balanced equations:$CO(NH_{2})_{2}_{(l)}=HNCO_{(l)}+NH_{3}_{(g)}$

$6HNO_{l}=C_{3}N_{3}(NH_{2})_{3}_{(l)}+3CO_{2}_{(g)}$

First we need to calculate the number of moles of urea that are used in the reaction, so:

molar mass of urea = $60.06\frac{g}{mol}*\frac{1kg}{1000g}=0.06006\frac{Kg}{mol}$

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$*\frac{1molofurea}{0.06006Kgofurea}=$2684 moles of urea

Now from the stoichiometry you have:

2684 moles of urea$*\frac{1molofHNCO}{1molurea}*\frac{1molofmelamine}{6molesofHNCO}$ = 447 moles of melamine

The molar mass of the melamine is $126.12\frac{g}{mol}$ so we have:

$447molesofmelamine*\frac{126.12g}{1molofmelamine}$ = 5637.64 g of melamine

Converting that mass of melamine to Kg:

5637.64 g of melamine *$\frac{1Kg}{1000g}$ = 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 = $\frac{76.5}{100}*56.38Kg$ = 43.13Kg of melamine