Question

The fertilizer ammonium sulfate [(NH₄)₂SO₄] is prepared by the reaction between ammonia (NH₃) and sulfuric acid: 2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(aq). How many kilograms of NH₃ are needed to produce 1.00 ⋅ 105 kg of (NH₄)₂SO₄?

Answer 1

To produce 100,000 kg of ammonium sulfate, approximately 25,755.69 kg of ammonia are needed, based on the molar masses and the stoichiometric relationship from the balanced chemical equation.

Step-by-step explanation:

To determine how many kilograms of ammonia (NH3) are needed to produce 100,000 kg (1.00 · 105 kg) of ammonium sulfate ((NH4)2SO4), we need to apply stoichiometry considerations from the balanced chemical equation:

2 NH3(g) + H2SO4(aq) → (NH4)2SO4(s).

This equation indicates that 2 moles of ammonia react with 1 mole of sulfuric acid to produce 1 mole of ammonium sulfate. To calculate the mass of ammonia, we will use the molar masses of ammonia and ammonium sulfate. The molar mass of ammonia is about 17.031 g/mol, and the molar mass of ammonium sulfate is about 132.14 g/mol.

Now, let's convert the mass of ammonium sulfate to moles using its molar mass and use the stoichiometric relationship to find the moles of ammonia needed:

1. Calculate the moles of (NH4)2SO4: 1.00 · 105 kg (NH4)2SO4 * (1,000,000 g/kg) / (132.14 g/mol) = 756,045.56 moles of (NH4)2SO4.
2. Use the mole ratio (2:1) of NH3 to (NH4)2SO4 to find moles of NH3: 756,045.56 moles (NH4)2SO4 * (2 moles NH3 / 1 mole (NH4)2SO4) = 1,512,091.12 moles NH3.
3. Convert moles of NH3 to kilograms: 1,512,091.12 moles NH3 * (17.031 g/mol) / (1,000,000 g/kg) ≈ 25,755.69 kg NH3.

Thus, approximately 25,755.69 kg of ammonia are needed to produce 100,000 kg of ammonium sulfate.