Question

Urea [(NH2)2CO] is prepared by reacting ammonia with with carbon dioxide
In the process 51g of NH3 are allowed to react with 176g of CO2

Answers

GPT-4o mini
To determine how much urea \((NH_2)_2CO\) can be produced from the reaction between ammonia \(NH_3\) and carbon dioxide \(CO_2\), we first need to write the balanced chemical equation for the reaction:

\[
2 NH_3 + CO_2 \rightarrow (NH_2)_2CO + H_2O
\]

From the balanced equation, we see that 2 moles of ammonia react with 1 mole of carbon dioxide to produce 1 mole of urea and 1 mole of water.

Next, we need to calculate the number of moles of \(NH_3\) and \(CO_2\) given the masses provided.

**Step 1: Calculate moles of \(NH_3\) and \(CO_2\)**

- Molar mass of \(NH_3\):
- \(N: 14.01 \, g/mol\)
- \(H: 1.008 \, g/mol\) (3 hydrogens)

\[
\text{Molar mass of } NH_3 = 14.01 + (3 \times 1.008) = 17.034 \, g/mol
\]

- Moles of \(NH_3\):
\[
\text{Moles of } NH_3 = \frac{51 \, g}{17.034 \, g/mol} \approx 3.00 \, mol
\]

- Molar mass of \(CO_2\):
- \(C: 12.01 \, g/mol\)
- \(O: 16.00 \, g/mol\) (2 oxygens)

\[
\text{Molar mass of } CO_2 = 12.01 + (2 \times 16.00) = 44.01 \, g/mol
\]

- Moles of \(CO_2\):
\[
\text{Moles of } CO_2 = \frac{176 \, g}{44.01 \, g/mol} \approx 4.00 \, mol
\]

**Step 2: Determine the limiting reactant**

According to the balanced equation, 2 moles of \(NH_3\) react with 1 mole of \(CO_2\).

- From 3 moles of \(NH_3\):
- Required moles of \(CO_2\) = \(\frac{3 \, mol \, NH_3}{2} = 1.5 \, mol \, CO_2\)

We have 4 moles of \(CO_2\) available, which is more than enough. Hence, \(NH_3\) is the limiting reactant.

**Step 3: Calculate the maximum amount of urea produced**

From the reaction, 2 moles of \(NH_3\) produce 1 mole of urea.

- Therefore, 3 moles of \(NH_3\) will produce:
\[
\text{Moles of urea} = \frac{3 \, mol \, NH_3}{2} = 1.50 \, mol \, urea
\]

**Step 4: Calculate the mass of urea produced**

- Molar mass of urea \((NH_2)_2CO\):
- \(N: 14.01 \, g/mol\) (2 nitrogens)
- \(H: 1.008 \, g/mol\) (4 hydrogens)
- \(C: 12.01 \, g/mol\)
- \(O: 16.00 \, g/mol\)

\[
\text{Molar mass of urea} = (2 \times 14.01) + (4 \times 1.008) + 12.01 + 16.00 = 60.06 \, g/mol
\]

- Mass of urea produced:
\[
\text{Mass of urea} = 1.50 \, mol \times 60.06 \, g/mol \approx 90.09 \, g
\]

In conclusion, the maximum amount of urea that can be prepared from the reaction of 51g \(NH_3\) with 176g \(CO_2\) is approximately **90.09 grams**.