When one mole of mixture of A2 and B2 allowed to react according to given reaction Calculate maximum energy released in This process
2A2(g) + 3B2(g) — 2A2B3(s) + 200 unit energy
A. 40 unit
B. 60 unit
C. 50 unit
D. 20 unit
Answers
Answer:
Explanation:
(i) 200 molecules of B will react with 200 atoms of A and 100 atoms of A will remain unreacted. Hence, B
2
is the limiting reagent.
(ii) 2 moles of A will react with 2 moles of B 2 and 1 mole of B 2 will remain unreacted. Thus, A is the limiting reagent.
(iii) 100 atoms of A will react with 100 molecules of B 2
. Thus, both reagents are present in stoichiometric amounts.
(iv) 2.5 mole of B 2 will react with 2.5 mole of A and 2.5 mole of A will remain unreacted. Hence, B 2 is the limiting reagent.
(v) 2.5 moles of A will react with 2.5 moles of B 2
and 2.5 moles of B 2
will remain unreacted. Thus, B 2
is the limiting reagent.
Answer:
The maximum energy released in the process When one mole of a mixture of A₂ and B₂ is allowed to react in the reaction is c)50 units.
Given:
The reaction given is
2A₂(g) + 3B₂(g) —> 2A₂B₃(s) + 200 unit energy.
To find:
The maximum energy is released in this process.
Solution:
1. First calculate the stoichiometric coefficients of the reactants and products:
A₂: 2 (reactant) and 0 (product)
B₂: 3 (reactant) and 1 (product)
A₂B₃: 0 (reactant) and 1 (product)
2. Calculate the mole ratio of A₂ to B₂:
A₂:B₂ = 2:3
3. Now calculate the maximum energy released in this reaction:
Energy released = (2 moles of A₂ * 200 units of energy) + (3 moles of B₂ * 200 units of energy)
= (2 * 200) + (3 * 200) = 50 units of energy
Therefore, the maximum energy released in this reaction is 50 units.
To know more about equilibrium, click here:
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