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:
Lo cual hace que la expresión de equilibrio se calcule como:
Y que en términos de la solubilidad molar, s, se resuelve como:
Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):
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<u>Answer:</u> The maximum amount of water that could be produced by the chemical reaction is 20.16 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of butane = 13 g
Molar mass of butane = 58.12 g/mol
Putting values in equation 1, we get:
Given mass of oxygen gas = 70.9 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of butane and oxygen gas follows:
By Stoichiometry of the reaction:
2 moles of butane reacts with 13 moles of oxygen gas
So, 0.224 moles of butane will react with = of oxygen gas
As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, butane is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of butane produces 10 moles of water
So, 0.224 moles of butane will produce = of water
Now, calculating the mass of water from equation 1, we get:
Molar mass of water = 18 g/mol
Moles of water = 1.12 moles
Putting values in equation 1, we get:
Hence, the maximum amount of water that could be produced by the chemical reaction is 20.16 grams