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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Answer:
Conduction, Convection, and Radiation .......
Answer:
2.73 is the equilibrium constant for the dissociation of 
gas at 840 degree Celsius.
Explanation:
Initial
0.600 atm 0
Equilibrium
(0.600 atm - p) 2p
Total pressure at equilibrium = P = 0.984 atm
P= 0.600 atm - p)+2p=0.984 atm
p = 0.384 atm
Partial pressure of the gas , 
= (0.600 atm - 0.384 atm)=0.216 atm
Partial pressure of the 
gas, 
= 2(0.384 atm)=0.768 atm
2.73 is the equilibrium constant for the dissociation of gas at 840 degree Celsius.
