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finlep [7]
3 years ago
14

500 gramos de un mineral con una riqueza en cinc del 65 % se hacen reaccionar con una disolucion de acido sulfurico de riqueza 9

6 % en peso y densidad 1823 kg/m3 . Calcula:
Chemistry
1 answer:
ryzh [129]3 years ago
5 0

Answer:

Ver explicación abajo

Explanation:

LA pregunta esta incompleta, pero logré conseguir un ejercicio muy parecido a este con datos similares, y la pregunta que falta como tal son estas:

<em>"a) La cantidad de sal producida. </em>

<em>b)  Moléculas de hidrógeno obtenidas a 25 ºC y 740 mm Hg. </em>

<em>c)  El volumen de la disolución de ácido necesario para la reacción."</em>

Asumiendo que estos son los datos faltantes, hay que ver como resolver por parte:

<u>a) Cantidad de sal producida.</u>

En este caso debemos plantear la reacción que se lleva a cabo. Es un mineral que tiene Zinc y reacciona con acido sulfurico, por tanto la reacción que se lleva a cabo en teoría es:

Zn + H₂SO₄ -----------> ZnSO₄ + H₂

Entonces, queremos saber la cantidad de ZnSO₄ que se formó.

Para eso, debemos calcular primero cuantos gramos de zinc hay originalmente en la muestra, que son los que reaccionaron para formar esta sal, y luego los moles para que por medio de estequiometría, calculemos los moles de la sal.

Para la masa de Zinc, sabemos que el mineral contiene 65% de zinc o 0,65 entonces:

mZn = 0,65 * 500 = 325 g

Calculamos los moles usando la masa molecular de Zinc que es 65,37 g/mol:

moles = 325 / 65,37 = 4.97 moles

Ahora bien, como la reacción de arriba está bien balanceada, podemos asuimr por estequiometría que la relación Zn / ZnSO₄ es 1:1, asi que los moles de Zn serán los mismos de la sal por tanto:

moles Zn = moles ZnSO₄ = 4,97 moles

Calculando la masa molecular de ZnSO₄:

MM ZnSO₄ = 65,37 + 32 + (16 * 4) = 161,37 g/mol

Finalmente la masa de la sal es:

m ZnSO₄ = 4,97 * 161,37

<h2>mZnSO₄ = 802,01 g</h2>

b) Moléculas de hidrógeno obtenidas

PAra este caso, ya tenemos los moles de Zn, y por estequiometría, todas las especies presentes están en relación 1:1, así que los moles de hidrógeno son los mismos de zinc. No es necesario el dato de temperatura y presión acá pues ya tenemos los moles.

Para conocer el número de moléculas, necesitamos el número de abogadro que es 6.02x10²³ por lo tanto las moléculas de hidrógeno:

Moléculas de hidrógeno = 6,02x10²³ * 4,97

<h2>Moléculas de hidrógeno = 2.99x10²⁴ moléculas de H₂</h2>

<u>c) Volúmen de ácido empleado</u>

Finalmente para el ácido, como ya conocemos los moles empleados, podemos calcular la masa del ácido usando su peso molecular:

MM H₂SO₄ = (2*1) + 32 + (4*16) = 98 g/mol

m H₂SO₄ = 4,97 * 98 = 487,06 g

Ahora que ya sabemos su masa, solo calculamos la masa realmente usada (pues su %p/p es 96%) y con la densidad calculamos el volumen:

d = m/v

v = m/d

m H₂SO₄ (puro) = 487,96 * 0,96 = 467,58 g

El volumen será:

V = 467,58 g / 1,823

<h2>V = 256,49 mL</h2>

Espero que esto te sirva, o te ayude como impulso a tu ejercicio real.

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