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Hunter-Best [27]
3 years ago
7

Determina el grado de pureza de un marmol (CaCO3), si al descomponerse 125 g del mismo se desprenden 20 litros de dióxido de car

bono medidos a 15ºC y 1 atm.
Chemistry
1 answer:
jarptica [38.1K]3 years ago
7 0

Answer:

67.8%

Explanation:

La reacción de descomposición del CaCO₃ es:

CaCO₃ → CO₂ + CaO

<em>Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.</em>

Usando la ley general de los gases, las moles de dioxido de carbono son:

PV = nRT.

<em>Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). </em>Reemplazando los valores en la ecuación:

PV / RT = n

1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles

Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.

La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:

0.846moles ₓ (100.087g / mol) = <em>84.7g de CaCO₃</em>

Así, la pureza del marmol es:

(84.7g de CaCO₃ / 125g) ₓ 100<em> = </em>

<h3>67.8%</h3>
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