Answer:
The percent yield of the reaction is 35 %
Explanation:
In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.
Let's verify the moles that were used in the reaction.
2.05 g . 1mol/ 32 g = 0.0640 mol
In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.
Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).
1atm . 0.550L = n . 0.082 . 295K
(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles
Percent yield of reaction = (Real yield / Theoretical yield) . 100
(0.0225 / 0.0640) . 100 = 35%
Answer:
[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka
Explanation:
The reaction of dissociation of the benzoic acid in water is given by the following equation:
C₆H₅-COOH + H₂O ⇄ C₆H₅-COO⁻ + H₃O⁺ (1)
The dissociation constant of an acid is the measure of the strength of an acid:
HA ⇄ A⁻ + H⁺ (2)
![K_{a} = \frac{[A^{-}][H^{+}]}{[HA]}](https://tex.z-dn.net/?f=%20K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%5BH%5E%7B%2B%7D%5D%7D%7B%5BHA%5D%7D%20)
(3)
<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA]. </em>
So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:
![K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}](https://tex.z-dn.net/?f=%20K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BC_%7B6%7DH_%7B5%7DCOO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DCOOH%5D%7D%20)
I hope it helps you!
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