Urea, (NH2)2CO is prepared by reacting ammonia with carbon dioxide.
The byproduct is water. If 136 g of ammonia are reacted with 264 g of carbon dioxide, which of the reactants in the limiting reactant?
The correct answer and explanation is :
To determine the limiting reactant, we follow these steps:
Step 1: Write the Balanced Chemical Equation
[
2NH_3 + CO_2 \rightarrow (NH_2)_2CO + H_2O
]
This equation shows that 2 moles of ammonia react with 1 mole of carbon dioxide to form 1 mole of urea and 1 mole of water.
Step 2: Convert Given Masses to Moles
- Molar masses:
- Ammonia (NH₃) = ( 14 + (3 \times 1) = 17 ) g/mol
- Carbon dioxide (CO₂) = ( 12 + (2 \times 16) = 44 ) g/mol
- Moles of ammonia:
[
\frac{136 \text{ g}}{17 \text{ g/mol}} = 8 \text{ moles of } NH_3
] - Moles of carbon dioxide:
[
\frac{264 \text{ g}}{44 \text{ g/mol}} = 6 \text{ moles of } CO_2
]
Step 3: Find the Limiting Reactant
From the balanced equation:
- 2 moles of NH₃ react with 1 mole of CO₂.
- 8 moles of NH₃ would require:
[
\frac{8}{2} = 4 \text{ moles of CO}_2
] - Since we have 6 moles of CO₂, this is more than enough.
- 6 moles of CO₂ would require:
[
6 \times 2 = 12 \text{ moles of NH}_3
] - But we only have 8 moles of NH₃, which is not enough.
Conclusion
Since ammonia runs out first, ammonia (NH₃) is the limiting reactant.
Here is an illustration of the reaction between ammonia (NH₃) and carbon dioxide (CO₂) to form urea ((NH₂)₂CO) and water (H₂O). The image visually represents the reaction process and highlights ammonia as the limiting reactant.