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.
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The reaction equation is:
Li + Br → LiBr
39 grams of Li = 39 / 7 = 5.57 moles of lithium
41.5 grams of Br = 41.5 / 80 = 0.52 mole of bromine
francium , in the Periodic table the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. making helium is the smallest element, and francium the largest.