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