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kupik [55]
3 years ago
9

Un mol de amoniaco Tiene una masa molar de 17 g y ocupa un volumen de 22.4 l qué volumen ocupa el 50 g amoniaco en condiciones n

ormales de temperatura y presión
Chemistry
1 answer:
mr_godi [17]3 years ago
4 0

Answer:

V  = 65.81 L

Explanation:

En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:

PV = nRT  (1)

Donde:

P: Presion (atm)

V: Volumen (L)

n: moles

R: constante de gases (0.082 L atm / mol K)

T: Temperatura (K)

De ahí, despejando el volumen tenemos:

V = nRT / P   (2)

Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:

n = 50 / 17 = 2.94 moles

Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:

V = 2.94 * 0.082 * 273 / 1

<h2>V = 65.81 L</h2>
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The reaction of equal molar amounts of benzene, C6H6, and chlorine, Cl2, carried out under special conditions, yields a gas and
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Answer : The molecular formula of a compound is, C_6H_5Cl_1

Solution : Given,

Mass of C = 64.03 g

Mass of H = 4.48 g

Mass of Cl = 31.49 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of Cl = 35.5 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{64.03g}{12g/mole}=5.34moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4.48g}{1g/mole}=4.48moles

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{31.49g}{35.5g/mole}=0.887moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{5.34}{0.887}=6.02\approx 6

For H = \frac{4.48}{0.887}=5.05\approx 5

For Cl = \frac{0.887}{0.887}=1

The ratio of C : H : Cl = 6 : 5 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_6H_5Cl_1

The empirical formula weight = 6(12) + 5(1) + 1(35.5) = 112.5 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{112.5}{112.5}=1

Molecular formula = (C_6H_5Cl_1)_n=(C_6H_5Cl_1)_1=C_6H_5Cl_1

Therefore, the molecular of the compound is, C_6H_5Cl_1

5 0
3 years ago
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