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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

Respuesta :

sebassandin

Answer:

[tex]n_{CO(NH_2)_2}=73.3molCO(NH_2)_2[/tex]

Explanation:

Hello.

In this case, given the two balanced chemical reactions:

[tex]CO(NH_2)_2 (l) \rightarrow HNCO(l) + NH_3(g)\\\\ 6HNCO(l) \rightarrow C_3N_3(NH_2)_3 (l) + 3CO_2(g)[/tex]

We first compute the moles of HNCO from the obtained 1.00 kg of melamine (molar mass 126 g/mol) by considering the 65 % yield:

[tex]m_{C_3N_3(NH_2)_3}^{theoretical}=\frac{1.00kg}{0.65}=1.54kg[/tex]

[tex]n_{HNCO}=1.54kgC_3N_3(NH_2)_3*\frac{1000g}{1kg}*\frac{1molC_3N_3(NH_2)_3}{126gC_3N_3(NH_2)_3} *\frac{6molHNCO}{1molC_3N_3(NH_2)_3} \\\\n_{HNCO}=73.3molHNCO[/tex]

Next, we compute the moles of urea in the first chemical reaction:

[tex]n_{CO(NH_2)_2}=73.3molHNCO*\frac{1molCO(NH_2)_2}{1molHNCO} \\\\n_{CO(NH_2)_2}=73.3molCO(NH_2)_2[/tex]

Best regards.

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