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(NH2)2 (l)  HNCO(l) + NH3(g) (balanced)
(2) HNCO(l)  C3N3(NH2)3 (l) + CO2(g) (unbalanced

Answer 1

Molar mass of melamine $=36+42+48=126\ g$

Moles of melamine in $1.00\ Kg$ of melamine $=\frac{1000}{126}=7.94$

The reaction $\ 6HNCO(l) \to C_3N_3(NH_2)_3 (l) + 3CO_2(g)$

$1$ mole of melamine requires $6$ moles of $HNCO.$

And given $65$ % yield of reaction,

Let $x$ moles of $HNCO$ are required to produce $7.94$ moles of melamine.

$\frac{65}{100}\times x\times \frac{1}{6}=7.94\implies x=73.29$

Now as per the reaction,

$CO(NH_2)_2 (l) \to HNCO(l) + NH_3(g)$

Moles of urea required be $y$ for production of $73.29$ moles of $HCNO$ given $100$ % yield .

Also, $1$ mole of urea is required for $1$ mole of $HCNO.$

So, $\frac{100}{100}\times y\times\frac {1}{1}=73.29\implies y=73.29$

So, moles of urea required are $73.29.$