Answer:
0.529
Explanation:
Let's consider the reaction A → Products
Since the units of the rate constant are s⁻1, this is a first-order reaction with respect to A.
We can find the concentration of A at a certain time t (![[A]_{t}](https://tex.z-dn.net/?f=%5BA%5D_%7Bt%7D)
) using the following expression.
![[A]_{t}=[A]_{0}.e^{-k\times t}](https://tex.z-dn.net/?f=%5BA%5D_%7Bt%7D%3D%5BA%5D_%7B0%7D.e%5E%7B-k%5Ctimes%20t%7D)
where,
[A]₀: initial concentration of A
k: rate constant
![[A]_{t}=0.548M.e^{-3.6\times 10^{-4}s^{-1}\times 99.2s }](https://tex.z-dn.net/?f=%5BA%5D_%7Bt%7D%3D0.548M.e%5E%7B-3.6%5Ctimes%2010%5E%7B-4%7Ds%5E%7B-1%7D%5Ctimes%2099.2s%20%7D)
![[A]_{t}=0.529 M](https://tex.z-dn.net/?f=%5BA%5D_%7Bt%7D%3D0.529%20M)
The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
The answer is Equivalence point. When the condition of [H+] = [OH-<span>] is reached, this state is called the </span><span>equivalence point.</span>
To much letters and numbers
Answer:
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