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motikmotik
3 years ago
15

La aspirina se prepara haciendo reaccionar ácido salicílico con exceso de anhídrido etanoico. En un experimento, 50.05 g de ácid

o salicílico se convirtieron en 55.45 g de aspirina. ¿Cuál fue el porcentaje de rendimiento?
Chemistry
1 answer:
Tanya [424]3 years ago
7 0

La aspirina se prepara haciendo reaccionar ácido salicílico con exceso de anhídrido etanoico. En un experimento, 50.05 g de ácido salicílico se convirtieron en 55.45 g de aspirina. ¿Cuál fue el porcentaje de rendimiento?

<em>In English:</em>

Aspirin is prepared by reacting salicylic acid with excess ethanoic anhydride. In one experiment, 50.05 g of salicylic acid was converted to 55.45 g of aspirin. What was the yield percentage?

Answer:

el rendimiento porcentual para la cantidad dada de ácido salicílico es 84.99 %

<em>In English:</em>

<em>the percent yield for the given amount of salicylic acid is </em><em>84.99%</em>

<em></em>

Explanation:

La ecuación química equilibrada para la reacción se puede escribir como:

C₇H₆O₃ + C₄H₆O₃    →    C₉H₈O₄ + HC₂H₃O₂

Para la reacción mostrada arriba; El reactivo limitante de la reacción es el ácido salicílico. Ahora; calcular el porcentaje de rendimiento; se espera que primero determinemos el rendimiento teórico de la reacción.

Entonces; la fórmula para calcular el porcentaje de rendimiento: \mathbf {= \frac{actual \ yield }{theoretical \ yield } *100 }  

El rendimiento teórico se determina de la siguiente manera:

50.05 g * 1 mol / 138.21 g / mol de C₇H₆O₃ * 1 mol de C₉H₈O₄ / 1 mol de C₇H₆O₃ * 180.157 g / mol de C₉H₈O₄ = 65.24 g de C₉H₈O₄

Porcentaje de rendimiento \mathbf {= \frac{55.45 }{65.24 } *100 }

Porcentaje de rendimiento = 84.99%

Por lo tanto, el porcentaje de rendimiento para la cantidad dada de ácido salicílico es 84.99%

<em>In English:</em>

<em>The balanced chemical eqaution for the reaction can be written as:</em>

<em>C₇H₆O₃ + C₄H₆O₃    →    C₉H₈O₄ + HC₂H₃O₂</em>

<em>For the reaction shown above;  The limiting reactant from the reaction is  salicylic acid. Now; to calculate the percentage yield ; we are expected to first determine the theoretical yield of the reaction. </em>

<em>So; the formula for calculating the percentage yield </em>\mathbf {= \frac{actual \ yield }{theoretical \ yield } *100 }<em>  </em>

<em />

<em>The theoretical yield is determined as follows:</em>

<em>50.05 g * 1 mol/ 138.21 g/mol of C₇H₆O₃ * 1 mol of C₉H₈O₄/ 1 mol of C₇H₆O₃ * 180.157 g/mol of C₉H₈O₄ = 65.24 g of C₉H₈O₄ is produced</em>

<em />

<em>Percentage yield </em>\mathbf {= \frac{55.45 }{65.24 } *100 }<em />

<em>Percentage yield = 84.99%</em>

<em />

<em>Thus, the percent yield for the given amount of salicylic acid is </em><em>84.99%</em>

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Answer:

9.1

Explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M

Step 4: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1

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3 years ago
375 mL of a 0.88 M potassium hydroxide solution is added to 496 mL of a 0.76 M cesium hydroxide solution. Calculate the pOH of t
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Answer:

pOH of resulting solution is 0.086

Explanation:

KOH and CsOH are monoacidic strong base

Number of moles of OH^{-} in 375 mL of 0.88 M of KOH = \frac{0.88\times 375}{1000}moles = 0.33 moles

Number of moles of OH^{-} in 496 mL of 0.76 M of CsOH = \frac{0.76\times 496}{1000}moles = 0.38 moles

Total volume of mixture = (375 + 496) mL = 871 mL

Total number of moles of OH^{-} in mixture = (0.33 + 0.38) moles = 0.71 moles

So, concentration of OH^{-} in mixture, [OH^{-}] = \frac{0.71}{871}\times 1000M=0.82M

Hence, pOH=-log[OH^{-}]=-log(0.82)=0.086

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2 years ago
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Answer:

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3 years ago
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If a buffer solution is 0.190 m in a weak acid (ka = 8.2 × 10-5) and 0.590 m in its conjugate base, what is the ph?\
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2 years ago
The three naturally occurring isotopes of potassium are 39K, 38.963707u; 40K, 39.963999u; and 41K.The percent natural abundances
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Answer:

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Explanation:

Step 1: Data given

The isotopes are:  

39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%

40K with an isotopic mass of 39.963999u

41K wit natural abundance of 6.7302 %

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Step 2: Calculate natural abundance of 40 K

100 % - 93.2581 % - 6.7302 %

100 % = 0.0117 %

Step 3: Calculate isotopic mass of 41K

39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302

39.098 = 36.33681 + 0.0046758 +  X * 2.067302

X = 40.9574 amu

The isotopic mass of 41K is 40.9574 amu

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