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Veseljchak [2.6K]
3 years ago
9

A partir de 140 g de n2 y una cantidad de h2 se obtuvo 153G de NH³. Cual es el % de rendimiento de la reacción química

Chemistry
1 answer:
DiKsa [7]3 years ago
4 0

Respuesta:

90.0 %

Explicación:

Paso 1: Escribir la ecuación química balanceada

N₂ + 3 H₂ ⇒ 2 NH₃

Paso 2: Calcular el rendimiento teórico de NH₃ a partir de 140 g de N₂

En la ecuación balanceada, participan de N₂: 1 mol × 28.01 g/mol = 28.01 g y de NH₃: 2 mol × 17.03 g/mol = 34.06 g.

140 g N₂ × 34.06 g NH₃ /28.01 g N₂ = 170 g NH₃

Paso 3: Calcular el rendimiento porcentual de NH₃

El rendimiento experimental de NH₃ es 153 g. Podemos calcular el rendimiento porcentual usando la siguiente fórmula.

R% = rendimiento experimental / rendimiento teórico × 100%

R% = 153 g / 170 g × 100% = 90.0 %

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Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂    

b) m = 5,043 g

c) % = 69,4 %

Explanation:

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4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂              

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.  

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles      

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De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles

m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

\% = \frac{R_{r}}{R_{T}}*100

<u>Donde</u>:

R_{r}: es el rendimiento real

R_{T}: es el rendimiento teórico

\% = \frac{3,5}{5,043}*100 = 69,4

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!        

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The reaction for photosynthesis producing glucose sugar and oxygen gas is:
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<u><em>1.5 grams of glucose is produced from 2.20 g of CO₂.</em></u>

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The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

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moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}

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Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

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