Question

Ammonium sulfate is used as a nitrogen and sulfur fertilizer. It is produced by reacting ammonia with sulfuric acid. Write the balanced equation for the reaction of gaseous ammonia with sulfuric acid solution. What volume (in liters) of ammonia at \(15^{\circ} \mathrm{C}\) and 1.15 atm is required to produce \(150.0 \mathrm{~g}\) of ammonium sulfate?

Short answer

46.17 liters of ammonia gas are required.

Step-by-step solution

Write the Chemical Reaction

The chemical reaction between ammonia (NH₃) and sulfuric acid (H₂SO₄) to produce ammonium sulfate ((NH₄)₂SO₄) is written as:\[2 \text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4\]This balanced equation shows that two moles of ammonia react with one mole of sulfuric acid to produce one mole of ammonium sulfate.

Calculate Moles of Ammonium Sulfate

First, we calculate the molar mass of ammonium sulfate: - \(\text{N}: 14.01 \times 2 = 28.02\)- \(\text{H}: 1.01 \times 8 = 8.08\)- \(\text{S}: 32.07 \times 1 = 32.07\)- \(\text{O}: 16.00 \times 4 = 64.00\)Thus, the molar mass of \((\text{NH}_4)_2\text{SO}_4\) is \(28.02 + 8.08 + 32.07 + 64.00 = 132.17 \text{ g/mol}\).Next, calculate the moles of ammonium sulfate required:\[\text{Moles of } (\text{NH}_4)_2\text{SO}_4 = \frac{150.0 \text{ g}}{132.17 \text{ g/mol}} = 1.135 \text{ mol}\]

Use Stoichiometry to Find Moles of Ammonia

From the balanced equation:\[2 \text{ moles of NH}_3 \rightarrow 1 \text{ mole of } (\text{NH}_4)_2\text{SO}_4\]To produce 1.135 moles of ammonium sulfate, ammonia needed (in moles):\[2 \times 1.135 = 2.270 \text{ moles of NH}_3\]

Use Ideal Gas Law to Find Volume of Ammonia

The ideal gas law is \( PV = nRT \).Where:- \( P = 1.15 \text{ atm} \)- \( n = 2.270 \text{ moles} \) of NH₃ calculated from Step 3.- \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \)- \( T = 15^{\circ}\text{C} + 273.15 = 288.15 \text{ K} \)Plug in the values to solve for \( V \):\[V = \frac{nRT}{P} = \frac{2.270 \times 0.0821 \times 288.15}{1.15} = 46.17 \text{ L}\]

Conclusion

The volume of ammonia gas required at \(15^{\circ} \text{C}\) and 1.15 atm to produce 150.0 g of ammonium sulfate is 46.17 liters.

Key concepts

Ammonium Sulfate Production

Ammonium sulfate is an important chemical compound widely used in agriculture as a nitrogen and sulfur fertilizer. Its production involves a chemical reaction between ammonia (NH₃) and sulfuric acid (H₂SO₄). During this reaction, ammonia, a basic gas, reacts with sulfuric acid, a strong acid, resulting in the formation of ammonium sulfate \((\text{NH}_4)_2\text{SO}_4\). This reaction can be represented by the balanced chemical equation: \[2 \text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 \] where two moles of ammonia react with one mole of sulfuric acid to produce one mole of ammonium sulfate. This balance ensures that all atoms in the reactants are accounted for in the products. Understanding this reaction is key to mastering stoichiometry and chemical reactions.

Ideal Gas Law

The Ideal Gas Law is a fundamental concept in chemistry that relates the pressure, volume, temperature, and moles of a gas. It is expressed by the equation: \[ PV = nRT \] where \( P \) stands for pressure, \( V \) is volume, \( n \) denotes moles of gas, \( R \) is the ideal gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\)), and \( T \) is the temperature in Kelvin. This law is crucial when performing calculations involving gases, such as determining the volume of gas needed or produced in a reaction. In the context of ammonium sulfate production, knowing the amount of ammonia gas required involves using the Ideal Gas Law to find the volume of gas when conditions such as temperature and pressure are specified. This helps to accurately measure required materials, crucial for industrial applications.

Chemical Reaction Balancing

Balancing chemical reactions is a fundamental skill in chemistry that ensures the conservation of mass in chemical equations. Each element must have the same number of atoms on both sides of the equation. In the reaction of ammonia with sulfuric acid to produce ammonium sulfate, the balancing is achieved as shown: \[ 2 \text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 \]Here, there are two nitrogen and six hydrogen atoms from the ammonia reacting with four additional hydrogen atoms and one sulfur atom from sulfuric acid, resulting in the combined formula of ammonium sulfate. Balancing equations requires practice but once mastered, it provides a solid foundation for predicting the outcomes of chemical reactions and their stoichiometric calculations.

Molecular Weight Calculation

Molecular weight calculations are necessary to convert masses of compounds into moles, an essential step in stoichiometry. The molecular weight of a compound is the sum of the atomic weights of all the atoms in its chemical formula.For example, ammonium sulfate \((\text{NH}_4)_2\text{SO}_4\) has a molecular weight calculated as follows:

• Nitrogen (N): \( 14.01 \times 2 = 28.02 \)
• Hydrogen (H): \( 1.01 \times 8 = 8.08 \)
• Sulfur (S): \( 32.07 \times 1 = 32.07 \)
• Oxygen (O): \( 16.00 \times 4 = 64.00 \)
• Total: \( 28.02 + 8.08 + 32.07 + 64.00 = 132.17 \text{ g/mol} \)

Knowing how to perform these calculations allows chemists to determine how many moles of a substance are present in a given mass, which is essential for calculating reactant or product quantities in chemical reactions.