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amm1812
3 years ago
14

During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that

the following side reaction (ii) occurs. (i) N2O4(l) 2 N2H4(l) 3 N2(g) 4 H2O(g) (ii) 2 N2O4(l) N2H4(l) 6 NO(g) 2 H2O(g) In one experiment 12.7 g of NO formed when 101.1 g of each reactant was used. What is the highest percent yield of N2 that can be expected
Chemistry
1 answer:
bonufazy [111]3 years ago
4 0

Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction:  2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

<em>N₂O₄ is the limiting reactant</em>

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ =  46.15 g of  N₂

In the side reaction, (6 * 30 g) of NO  is produced  from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ =  40.23 g of  N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%

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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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