Question

. What mass of (NH),SO, must be mixed with 35.0 kg of KNO3 to produce a fertilizer that has 15.1 mass % nitrogen?

Answer 1

Mass of (NH₄)₂SO₄ = 7.20 kg

Explanation

mass of KNO₃ = 35.0 kg

moles KNO₃ = (mass KNO₃) / (molar mass KNO₃)

moles KNO₃ = (35.0 kg) / (101.1032 g/mol)

moles KNO₃ = 0.34618 kmol

moles N₂ = moles KNO₃

moles N₂ = 0.34618 kmol

mass N₂ = (moles N₂) * (molar mass N₂)

mass N₂ = (0.34618 kmol) * (14.0 g/mol)

mass N₂ = 4.846533 kg

Let mass (NH₄)₂SO₄ = x kg

moles (NH₄)₂SO₄ = (mass (NH₄)₂SO₄) / (molar mass (NH₄)₂SO₄)

moles (NH₄)₂SO₄ = (x kg) / (132.1395 g/mol)

moles (NH₄)₂SO₄ = x/132.1395 kmol

moles N₂ = 2 * (moles (NH₄)₂SO₄)

moles N₂ = 2 * (x/132.1395 kmol)

moles N₂ = 2x/132.1395 kmol

mass N₂ = (moles N₂) * (molar mass N₂)

mass N₂ = (2x/132.1395 kmol) * (14.0 g/mol)

mass N₂ = 28x/132.1395 kg

Total mass N₂ = 4.846533 kg + 28x/132.1395 kg

Total mass fertilizer = (mass KNO₃) + (mass (NH₄)₂SO₄)

Total mass fertilizer = (35.0 kg) + (x kg)

Total mass fertilizer = (35.0 + x) kg

mass % nitrogen = (Total mass N₂ / Total mass fertilizer) * 100

15.1 = [(4.846533 kg + 28x/132.1395 kg) / (35.0 + x) kg)] * 100

0.151 = (4.846533 + 28x/132.1395) / (35.0 + x)

0.151 * (35.0 + x) = 4.846533 + 28x/132.1395

Solving for x, x = 7.20 kg

mass (NH₄)₂SO₄ = x kg = 7.20 kg