Question

1. urea (h₂nconh₂, m = 60.055 g/mol) can be produced by the following reaction: co₂(g) + 2 nh₃(g) → h₂nconh₂(s) + h₂o(g) what is the maximum mass of urea (in g) that can be produced if 5.06 l of nh₃(g) at a pressure of 2.39 bar is mixed with 0.255 moles of co₂(g) at 123°c?

Explanation:

Step1: Calculate moles of \(NH_3\) using ideal gas law

The ideal gas law is \(PV = nRT\).
Given \(P = 2.39\ bar=2.39\times10^{5}\ Pa\), \(V = 5.06\ L = 5.06\times10^{- 3}\ m^{3}\), \(T=(123 + 273.15)K=396.15\ K\), \(R = 8.314\ J/(mol\cdot K)\)

$$n=\frac{PV}{RT}=\frac{2.39\times10^{5}\times5.06\times10^{-3}}{8.314\times396.15}$$

$$n=\frac{1219.34}{3304.13}\approx0.369\ mol$$

Step2: Determine the limiting reactant

From the reaction \(CO_{2}(g)+2NH_{3}(g)\to H_{2}NCONH_{2}(s)+H_{2}O(g)\)
The mole ratio of \(CO_{2}\) to \(NH_{3}\) is \(1:2\)
For \(n(CO_{2}) = 0.255\ mol\), the moles of \(NH_{3}\) required is \(2\times0.255 = 0.51\ mol\)
For \(n(NH_{3})=0.369\ mol\), the moles of \(CO_{2}\) required is \(\frac{0.369}{2}=0.1845\ mol\)
Since \(0.369\ mol\) of \(NH_{3}\) is less than the amount required for \(0.255\ mol\) of \(CO_{2}\) (\(0.51\ mol\)), \(NH_{3}\) is the limiting reactant.

Step3: Calculate moles of urea

From the reaction, \(2\ mol\) of \(NH_{3}\) produces \(1\ mol\) of urea.
If \(n(NH_{3}) = 0.369\ mol\), then \(n(urea)=\frac{0.369}{2}=0.1845\ mol\)

Step4: Calculate mass of urea

Given \(M(urea)=60.055\ g/mol\)

$$m = n\times M=0.1845\times60.055$$

$$m\approx11.1\ g$$

Answer:

\(11.1\ g\)