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Mama L [17]
2 years ago
7

The fertilizer ammonium sulfate, (NH4)2SO4, is prepared by the reaction between ammonia (NH3) and sulfuric acid:

Chemistry
1 answer:
kumpel [21]2 years ago
4 0

The mass of ammonia required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄ is 6.18 * 10⁴ Kg of ammonia.

<h3>What mass in kilograms of ammonia are required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄?</h3>

The mass of ammonia required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄ is determined from the mole ratio of the reaction.

The mole ratio of the reaction is obtained from the balanced equation of the reaction given below:

  • 2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(aq)

Mole ratio of NH₃ and (NH₄)₂SO₄ is 2: 1

Mass of 2 moles of ammonia = 2 * 17 = 34 g

Mass of 1 mole of (NH₄)₂SO₄ = 132 g

Mass of ammonia required = 34/132 *  2.40 × 10⁵ kg

Mass of ammonia required = 6.18 * 10⁴ Kg of ammonia.

In conclusion, the mole ratio is used to determine the mass of ammonia required.

Learn more about mole ratio at: brainly.com/question/19099163

#SPJ1

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