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)

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  • 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

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