Urea [(NH2)2CO] is prepared by reacting ammonia with with carbon dioxide

Question

In the process 51g of NH3 are allowed to react with 176g of CO2
Which of the two reactant is the limiting reagent
Calculate the mass of (NH2)2CO formed
How much of the excess reagent (in grams)is left at the end of the reaction

Answer 1

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

1. Moles of NH3: \[ \text{Molar mass of NH}_3 = 14.01 , (\text{N}) + 3 \times 1.008 , (\text{H}) = 17.034 , \text{g/mol} \] \[ \text{Moles of NH}_3 = \frac{51 , \text{g}}{17.034 , \text{g/mol}} \approx 3.00 , \text{mol} \]

2. 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