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icang [17]
3 years ago
6

Which of the following conversion factors would be used to calculate the number of moles of Cl2 produced if 11 moles of HCl were

present in the following reaction:______
4HCl O2 --> 2Cl2 2H2O
Chemistry
1 answer:
MissTica3 years ago
6 0

<u>Answer:</u> The moles of chlorine gas produced is 5.5 moles

<u>Explanation:</u>

We are given:

Moles of HCl = 11 moles

For the given chemical reaction:

4HCl+O2\rightarrow 2Cl_2+2H_2O

By Stoichiometry of the reaction:

4 moles of HCl produces 2 moles of chlorine gas

So, 11 moles of HCl will be produced from \frac{2}{4}\times 11=5.5mol of chlorine gas

Hence, the moles of chlorine gas produced is 5.5 moles

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Answer:

The amount of water vapor in a given volume of air is humidity.The amount of water vapor in the air at any one time depends on temperature . Warmer air holds more water than cooler air. Clouds form when humidity is low and thetemperature is high . Relative humidity is defined as the amount of water vapor in the air divided by the amount that would have to be present in the same air to form a cloud or condense on a surface. Dew is related to the humidity of the previous day. The air temperature at which water vapor in the air condenses onto cool surfaces is the dew point.

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Rocks beneath the surface are forced toward the________.
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the mantle

Explanation:

7 0
2 years ago
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1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en
Maslowich

Answer:

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

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

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      

n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles

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!        

5 0
3 years ago
Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles
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Answer:

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Explanation:

This the unbalanced reaction

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The balanced reaction:

4NH₃  +  3O₂  →  2N₂  +  6H₂O

4 mol of ammonia

3 mol of oxygen

2 mol of nitrogen

6 mol of water

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