A sample of an unknown metal has a mass of 120.4 g. As the sample cools from 90.5°C to 25.7°C, it releases 7020 J of energy. Usi
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Answer:
H₂SO₄ will be the limiting reagent.
Explanation:
The balanced reaction is:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction).
You can use a simple rule of three as follows: if by stoichiometry 2 moles of Al(OH)₃ reacts with 3 moles of H₂SO₄, how much moles of H₂SO₄ will be needed if 0.4 moles of Al(OH)₃ react?
moles of H₂SO₄= 0.6 moles
But 0.6 moles of H₂SO₄ are not available, 0.40 moles are available. Since you have less moles than you need to react with 0.4 moles of Al(OH)₃, H₂SO₄ will be the limiting reagent.
Explanation:
Entropy means the amount of randomness present within the molecules of the body of a substance.
Relation between entropy and microstate is as follows.
S =
where, S = entropy
= Boltzmann constant
= number of microstates
This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.
Also, let us assume that = 1, and
= 0.833
Therefore, change in entropy will be calculated as follows.
=
=
=
or, =
Thus, we can conclude that the entropy change for a particle in the given system is J/K particle.