The theoretical yield is the
yield after a certain reaction occurs. So if you are given 17.1 g of H2 and an
excess of N2, you should know the balanced equation.
N2 + 3H2 → 2NH3
Molar mass of H2 = 2.02 g/mol
Molar mass of NH3 = 17 g/mol
(17.1 g of H2)(1 mol H2/2.02g H2)(2
mol NH3/3mol H2)(17g NH3/1mol NH3) = 95.94g NH3
Answer:
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)
Explanation:
Which ONE of the following is an oxidation–reduction reaction?
A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.
B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.
C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.
D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.
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
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A. It’s known as the “universal solvent”