Question

Phosgene and ammonia gases can react to produce urea and ammonium chloride solids according to the following chemical equation: \(\mathrm{COCl}_{2}(g)+4 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(s)+2 \mathrm{NH}_{4} \mathrm{Cl}(s)\) Determine the mass of each product formed when \(52.68 \mathrm{~g}\) \(\mathrm{COCl}_{2}(g)\) and \(35.50 \mathrm{~g} \mathrm{NH}_{3}(g)\) are combined. Which reactant is consumed completely? How much of the other reactant remains when the reaction is complete?

Short answer

NH₃ is the limiting reactant; 31.29g of urea and 55.76g of NH₄Cl form; 1.09g of COCl₂ remains.

Step-by-step solution

Calculate Molar Masses

To begin, calculate the molar mass of each reactant and product involved in the reaction:- \(\mathrm{COCl}_2\): \(12.01 + 2(35.45) + 16.00 = 98.91 \, \mathrm{g/mol}\)- \(\mathrm{NH}_3\): \(14.01 + 3(1.01) = 17.04 \, \mathrm{g/mol}\)- \(\mathrm{CO(NH}_2)_2\): \(12.01 + 2(14.01) + 4(1.01) + 16.00 = 60.06 \, \mathrm{g/mol}\)- \(\mathrm{NH}_4\mathrm{Cl}\): \(14.01 + 4(1.01) + 35.45 = 53.50 \, \mathrm{g/mol}\)

Determine Moles of Reactants

Calculate the number of moles of each reactant:- Moles of \(\mathrm{COCl}_2\): \(\frac{52.68\, \mathrm{g}}{98.91\, \mathrm{g/mol}} = 0.532\, \text{moles}\)- Moles of \(\mathrm{NH}_3\): \(\frac{35.50\, \mathrm{g}}{17.04\, \mathrm{g/mol}} = 2.082\, \text{moles}\)

Identify the Limiting Reactant

According to the balanced equation, 1 mole of \(\mathrm{COCl}_2\) reacts with 4 moles of \(\mathrm{NH}_3\). Calculate the amount of \(\mathrm{NH}_3\) needed for 0.532 moles of \(\mathrm{COCl}_2\): \(0.532 \times 4 = 2.128\, \text{moles}\). Since only 2.082 moles of \(\mathrm{NH}_3\) are available, \(\mathrm{NH}_3\) is the limiting reactant.

Calculate Amount of Products Formed

Each reaction produces 1 mole of \(\mathrm{CO(NH}_2)_2\) for every reaction between \(\mathrm{COCl}_2\) and \(\mathrm{NH}_3\), and 2 moles of \(\mathrm{NH}_4\mathrm{Cl}\). With \(\mathrm{NH}_3\) as the limiting reactant, use its mole amount:- \(\mathrm{CO(NH}_2)_2\): \(\frac{2.082}{4} \approx 0.521\, \text{moles of product}\)- \(\mathrm{NH}_4\mathrm{Cl}\): \((2) \times 0.521 \approx 1.042\, \text{moles of product}\)

Convert Moles of Products to Mass

Convert the moles of each product to grams:- Mass of \(\mathrm{CO(NH}_2)_2\): \(0.521 \text{ moles} \times 60.06\, \mathrm{g/mol} \approx 31.29\, \mathrm{g}\)- Mass of \(\mathrm{NH}_4\mathrm{Cl}\): \(1.042 \text{ moles} \times 53.50\, \mathrm{g/mol} \approx 55.76\, \mathrm{g}\)

Calculate Remaining Reactant

Determine the remaining moles of \(\mathrm{COCl}_2\) after the reaction:- Moles used: \(0.521\) (as each full reaction requires 1 mole of \(\mathrm{COCl}_2\))- Moles remaining: \(0.532 - 0.521 = 0.011\, \text{moles}\)Convert remaining moles of \(\mathrm{COCl}_2\) to grams:- Remaining \(\mathrm{COCl}_2\): \(0.011 \times 98.91\, \mathrm{g/mol} \approx 1.09\, \mathrm{g}\)

Key concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first, dictating the maximum amount of product that can be formed. In our example, we need to determine which reactant will run out first when phosgene (\(\text{COCl}_{2}\)) reacts with ammonia (\(\text{NH}_{3}\)). To find it, compare the mole ratio from the balanced chemical equation:- 1 mole of \(\text{COCl}_{2}\) requires 4 moles of \(\text{NH}_{3}\). - The calculation in the exercise showed 0.532 moles of \(\text{COCl}_{2}\) would require \(0.532 \times 4 = 2.128\) moles of \(\text{NH}_{3}\).- However, only \(2.082\) moles of \(\text{NH}_{3}\) are available.Since \(\text{NH}_{3}\) is insufficient for the reaction with all the phosgene, it becomes the limiting reactant.

Molar Mass

Molar mass is the mass of one mole of a substance, critical for converting grams to moles or vice versa. It's calculated by summing the atomic masses of each element in a compound, available on the periodic table.Here's how we calculate it for each compound:

• For \(\text{COCl}_{2}\), add the atomic masses: Carbon (C) \(12.01\), Oxygen (O) \(16.00\), and two Chlorines (Cl) \(35.45 \times 2 = 70.90\). Thus, \(\text{COCl}_{2}: 12.01 + 16.00 + 70.90 = 98.91\, \text{g/mol}\).
• For \(\text{NH}_{3}\), one Nitrogen (N) \(14.01\) and three Hydrogens (H) \(1.01 \times 3 = 3.03\). So, \(\text{NH}_{3}: 14.01 + 3.03 = 17.04\, \text{g/mol}\).

Understanding molar mass allows us to precisely measure and plan for chemical reactions.

Chemical Reaction

A chemical reaction involves the rearrangement of atoms to transform reactants into products. In our exercise, the defined chemical reaction is: \[\text{COCl}_{2}(g) + 4\, \text{NH}_{3}(g) \rightarrow \text{CO(NH}_{2})_{2}(s) + 2\, \text{NH}_{4}\text{Cl}(s)\]The left side lists the reactants: phosgene and ammonia gases. The right side shows the products: urea and ammonium chloride solids.Key aspects of chemical reactions include:

• **Observing the Law of Conservation of Mass:** No atoms are lost or gained - they're simply redistributed to form new compounds.
• **Reactants to Products:** The initial substances (reactants) interact, breaking old bonds and forming new ones to create different substances (products).

Product Formation

The reaction between phosgene and ammonia leads to the formation of two products, urea and ammonium chloride. The amount of product depends directly on the moles of the limiting reactant, ammonia in this case.According to the balanced equation:- 1 mole of \(\text{COCl}_{2}\) with 4 moles of \(\text{NH}_{3}\) will form 1 mole of \(\text{CO(NH}_{2})_{2}\) and 2 moles of \(\text{NH}_{4}\text{Cl}\).With \(2.082\) moles of \(\text{NH}_{3}\), we achieve:

• \(\text{CO(NH}_{2})_{2}: \frac{2.082}{4} = 0.521\, \text{moles}\)
• \(\text{NH}_{4}\text{Cl}: 2 \times 0.521 = 1.042\, \text{moles}\)

Finally, convert moles to grams to find the mass of each product. This conversion is essential to understand how much of each substance is produced in real-world terms.