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densk [106]
3 years ago
10

2 Cu + Cl2 ----> 2 Cuci

Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
4 0

Answer:

320 g.

Explanation:

Hello!

In this case, according to the balanced chemical reaction, we can compute the grams of copper I chloride produced by each reactant, as shown below:

m_{CuCl}^{by\ Cu}=3.23molCu*\frac{2molCuCl}{2molCu}*\frac{99.0gCuCl}{1molCuCl}  =320gCuCl\\\\m_{CuCl}^{by\ Cl_2}=1.64molCl_2*\frac{2molCuCl}{1molCl_2}*\frac{99.0gCuCl}{1molCuCl}  =325gCuCl

Thus, since copper produces the fewest grams of CuCl, we infer it is the limiting reactant, therefore the correct mass of copper I chloride is 320 g.

Best regards!

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2 years ago
At 500K, the equilibrium constant for the reaction N2O4D2NO2 is 1.5 x 10^3. What is the equilibrium constant for the reaction: 6
Veronika [31]

Answer:

K = 2.96x10⁻¹⁰

Explanation:

Based on the initial reaction:

N2O4 ⇄ 2NO2; K = 1.5x10³

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3 times this reaction:

3N2O4 ⇄ 6NO2; K = (1.5x10³)³ =3.375x10⁹

The inverse reaction has a K of:

6NO2 ⇄ 3N2O4 K = 1/3.375x10⁹;

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3 0
3 years ago
What product is obtaned when carbon is
kirza4 [7]

Answer:

carbon dioxide

Explanation:

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Carbon, as graphite, burns to form gaseous carbon (IV) oxide (carbon dioxide), CO2. ... When the air or oxygen supply is restricted, incomplete combustion to carbon monoxide, CO, occurs. 2C(s) + O2(g) → 2CO(g) This reaction is important. When one mole of carbon is exposed to some energy in the presence of one mole of oxygen gas, one mole of carbon dioxide gas is produced. This reaction is a combustion reaction.

6 0
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El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si
Alchen [17]

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)

Lo cual hace que la expresión de equilibrio se calcule como:

Ksp=[Pb^{2+}][Cl^-]^2

Y que en términos de la solubilidad molar, s, se resuelve como:

1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L

¡Saludos!

7 0
2 years ago
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