Answer:
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Explanation:
For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.
For a hypothetical reaction:
xA + yB ⇄ zC
The equilibrium constant is :
![Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D%7D%7B%5BC%5D%5E%7Bz%7D%20%7D)
The given reaction involves the decomposition of H2O into H2 and O2
The equilibrium constant is expressed as :
![Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Since Keq = 5.31*10^-10
Answer:
Percent yield = 90.5%
Explanation:
Given data:
Mass of carbon dioxide = 500 g
Mass of water = excess
Actual yield of carbonic acid = 640 g
Percent yield = ?
Solution:
Balanced chemical equation:
CO₂ + H₂O → H₂CO₃
Number of moles of carbon dioxide
Number of moles = Mass / molar mass
Number of moles = 500 g/ 44 g/mol
Number of moles = 11.4 mol
Now we will compare the moles of H₂CO₃ with CO₂.
CO₂ : H₂CO₃
1 : 1
11.4 : 11.4
Mass of carbonic acid:
Mass = number of moles × molar mass
Mass = 11.4 mol × 62.03 g/mol
Mass = 707.14 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 640 g/ 707.14 g × 100
Percent yield = 90.5%
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