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Anuta_ua [19.1K]
3 years ago
12

What is distancewho knows pls ASAP​

Chemistry
1 answer:
notsponge [240]3 years ago
6 0

Answer: an amount of space between two things or people.

Explanation: hope this helps? :)

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La aspirina se prepara haciendo reaccionar ácido salicílico con exceso de anhídrido etanoico. En un experimento, 50.05 g de ácid
Tanya [424]

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>

7 0
3 years ago
onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
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