Anonymous
Anonymous asked in Science & MathematicsChemistry · 4 weeks ago

# asap help ?

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)

Relevance
• 4 weeks ago

Moles C3N3(NH2)3 = 126.12 g/mol

moles melamine desired = 1000 g / 126.12 g/mol = 7.93 moles

Balanced second equation:

6 HNCO(l) --> C3N3(NH2)3 (l) + 3 CO2(g)

With 100% yield, moles HNCO required:

7.93 mol C3N3(NH2)3 X (6 mol HNCO/ 1 mol C3N3(NH2)3) = 47.6 mol HNCO

With the second reaction having only a 65% yield, moles HNCO required:

47.6 / 0.65 = 73.2 mol HNCO

Moles and mass urea required:

73.2 mol HNCO X (1 mol urea/1 mol HNCO) = 73.2 mol urea

Mass urea = 73.2 mol X 60.06 g/mol = 4400 g = 4.40 kg urea