To determine the limiting reagent in the reaction between ammonia (NH3) and carbon dioxide (CO2) to form urea \((NH_2)_2CO\), we first need to write the balanced chemical equation for the reaction:
\[ 2 , \text{NH}_3 + \text{CO}_2 \rightarrow (NH_2)_2CO + \text{H}_2O \]
From the balanced equation, we can see that 2 moles of NH3 react with 1 mole of CO2 to produce 1 mole of urea.
Step 1: Calculate moles of each reactant
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Moles of NH3: \[ \text{Molar mass of NH}_3 = 14.01 , (\text{N}) + 3 \times 1.008 , (\text{H}) = 17.034 , \text{g/mol} \] \[ \text{Moles of NH}_3 = \frac{51 , \text{g}}{17.034 , \text{g/mol}} \approx 3.00 , \text{mol} \]
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Moles of CO2: \[ \text{Molar mass of CO}_2 = 12.01 , (\text{C}) + 2 \times 16.00 , (\text{O}) = 44.01 , \text{g/mol} \] \[ \text{Moles of CO}_2 = \frac{176 , \text{g}}{44.01 , \text{g/mol}} \approx 4.00 , \text{mol} \]
Step 2: Determine the limiting reagent
From the balanced equation, the stoichiometry is 2 moles of NH3 for every 1 mole of CO2. Thus, we need:
- 2 moles of NH3 for every 1 mole of CO2.
We have 3.00 moles of NH3: \[ \text{Required moles of CO}_2 = \frac{3.00 , \text{mol NH}_3}{2} = 1.50 , \text{mol CO}_2 \]
Since we have 4.00 moles of CO2 available, and we only need 1.50 moles, NH3 is the limiting reagent.
Step 3: Calculate the mass of urea formed
According to the balanced equation:
- 2 moles of NH3 produce 1 mole of urea.
Therefore, the moles of urea produced: \[ \text{Moles of urea} = \frac{3.00 , \text{mol NH}_3}{2} = 1.50 , \text{mol (NH}_2)_2CO \]
Now, calculate the mass of urea: \[ \text{Molar mass of (NH}_2)_2CO = 2 \times 14.01 , (\text{N}) + 4 \times 1.008 , (\text{H}) + 12.01 , (\text{C}) + 16.00 , (\text{O}) = 60.06 , \text{g/mol} \] \[ \text{Mass of urea} = 1.50 , \text{mol} \times 60.06 , \text{g/mol} \approx 90.09 , \text{g} \]
Step 4: Calculate the excess reagent left
Since NH3 is the limiting reagent, we need to check how much CO2 has been consumed: \[ \text{Moles of CO}_2 \text{ consumed} = \frac{3.00 , \text{mol NH}_3}{2} = 1.50 , \text{mol CO}_2 \]
Remaining moles of CO2: \[ \text{Remaining moles of CO}_2 = 4.00 , \text{mol CO}_2 - 1.50 , \text{mol CO}_2 = 2.50 , \text{mol CO}_2 \]
Now calculate the mass of remaining CO2: \[ \text{Mass of CO}_2 = 2.50 , \text{mol} \times 44.01 , \text{g/mol} \approx 110.03 , \text{g} \]
Summary of results
- Limiting reagent: NH3
- Mass of urea produced: 90.09 g
- Mass of excess CO2 remaining: 110.03 g