Question

Melamine (C3N3(NH2)3) is a component of many adhesives and resins and is manufactured in a two-step

process from urea (CO(NH2)2) as the sole starting material. How many moles of urea would be required if we want to collect 1.00 kg of melamine and if the first step in the process is 100% yield, but the second step is only
65% yield?
(1) CO(NH)2 (1) ► HNCO(1) + NH3(g) (balanced)
(2) HNCO(1) → C3N3(NH)3 (1) + CO2(g) (unbalanced)​

Answer 1

43.13Kg of melamine

Step-by-step explanation:

The problem gives you the mass of urea and two balanced equations:

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

molar mass of urea =

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 urea2684 moles of urea

Now from the stoichiometry you have:

2684 moles of urea = 447 moles of melamine

The molar mass of the melamine is so we have:

= 5637.64 g of melamine

Converting that mass of melamine to Kg:

5637.64 g of melamine * = 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 = = 43.13Kg of melamine