Using the van der Waals equation, calculate the pressure (in atmospheres) exerte

Question

Using the van der Waals equation, calculate the pressure (in atmospheres) exerted by 44.90 g of CH4 at 75.51 °C in a 1.00 L container. The van der Waals constants for CH4 are a = 2.250 L2*atm/mol2 and b = 0.04280 L/mol.

and

A sample contained a mixture of CdS and ZnS. The amount of CdS was determined by reacting 7.600 g of the sample with an excess of HCl to release H2S from each compound;

CdS(s) + 2 HCl(aq) CdCl2 + H2S(g)
ZnS(s) + 2 HCl(aq) ZnCl2 + H2S(g)

If the sample reacted completely and produced 1750 mL of hydrogen sulfide at 25.99 °C and 756.7 mm Hg, what was the mass percentage of CdS in the mixture?

Explanation / Answer

(1)

CH4 moles = mass / molar mass = 44.90 / 16 =2.81

T = 75.51 + 273 = 348.5 K

V = 1 L

a = 2.250 L2*atm/mol2

b = 0.04280 L/mol.

vander wall equation

(P + n^2 a / V^2) (V-nb) = n R T

(P + 2.81^2 x 2.250 / 1^1) (1 -2.81 x 0.04280) = 2.81 x 0.0821 x 348.5

(P + 2.81^2 x 2.250 / 1^1) (1 -2.81 x 0.04280) = 80.40

(P + 2.81^2 x 2.250 / 1^1) = 91.39

P +17.77 = 91.39

P = 73.62 atm

pressure = 73.62 atm

(2)

V = 1750 ml = 1.750 L

T = 273 + 25.99 = 298.99 K

P = 756.7 mmHg = 756.7 / 760 = 0.996 atm

R = 0.0821 L -atm /mol K

P V = nRT

0.996 x 1.750 = n x 0.0821 x 298.99

n = 0.07

moles = mass /molar mass

mass = moles x molar mass

         = 0.07 x 144.48

        = 0.000484 g

mass percent of CdS = mass of CdS / sample mass

                                   = 0.00484 / 7.60

                                  = 0.00637 %

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