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ladessa [460]
3 years ago
15

Two volatile liquids A (P A pure= 165 Torr) and B (PB pure= 85.1 Torr) are confined in a piston/cylinder assembly. Initially onl

y the liquid phase is present. As the external pressure is reduced, vapor is first observed at a total pressure of 110 Torr. Calculate the mole fraction of component A in the solution (XA) and the mole fraction of component A in the vapor (YA).
Chemistry
1 answer:
Assoli18 [71]3 years ago
6 0

Answer:

y_A=0.467\\x_A=0.312

Explanation:

Hello,

In this case, one could use the following equations to compute the required mole fractions (fugacity equality for both A and B):

p_A^{vap}x_A=y_AP\\p_B^{vap}(1-x_A)=(1-y_A)P\\p_B^{vap}-x_Ap_B^{vap}=P-y_AP\\y_AP-x_Ap_B^{vap}=P-p_B^{vap}

Therefore, the 2x2 system of equations turns out:

\left \{ {{165x_A-110y_A=0} \atop {110y_A-85.1x_A=24.9}} \right.\\ \\y_A=0.467\\x_A=0.312

Best regards.

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Explanation:

Equation of the reaction:

2KMnO4(aq) + 16HCl(aq) --> 2MnCl2(aq) + 2KCl(aq) + 5Cl2(g) + 8H2O(aq)

To calculate the limiting reagent, we need to calculate the number of moles of the reactants :

KMnO4:

Molar mass = (39 + 55 + (16*4))

= 158 g/mol

Number of moles = mass/molar mass

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Molar mass = 1 + 35.5

= 36.5 g/mol

Number of moles = 526.64/36.5

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The limiting reagent :

14.428 moles of HCl * 2 moles of KMnO4/16moles of HCl

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Mass of the products:

KCl:

2 moles of KMnO4 will produce 2 moles of KCl

Moles of KCl = 1 * 1.4506 mol

= 1.4506 mol

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Mass of KCl = 74.5 * 1.4506

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MnCl2:

2 moles of KMnO4 will produce 2 moles of MnCl2

Number of moles of MnCl2 = 1 * 1.4506

= 1.4506 mol

Molar mass = 55 + (35.5*2)

= 126 g/mol

Mass of MnCl2= 1.4506 * 126

= 182.78 g

Cl2:

2 moles of KMnO4 will produce 5 moles of Cl2

Number of moles of Cl2 = 5/2 * 1.4506

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Molar mass of Cl2 = 35.5*2

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Mass of Cl2 = 71* 3.6265

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H2O:

2 moles of KMnO4 will produce 8 moles of H2O

Number of moles of H2O = 8/2 * 1.4506

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Molar mass of H2O =( 1*2) + 16

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Mass of H2O = 18*5.80

= 104.44 g

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3 years ago
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