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A gas mixture contains each of the following gases at the indicated partial pressures: , 221O2torr , N2,131 torr; and He 131 torr
What mass of each gas is present in a 1.30 - L sample of this mixture at 25.0 degrees C?

Respuesta :

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 221 torr + 131 torr + 131 torr = 483 torr total 
n = PV / RT = (483 torr) x (1.30 L) / ((62.36367 L Torr/K mol) x (25.0 + 273.15 K)) = 0.033769 mol gases total 
(0.033769 mol) x ( 221 torr N2/ 483 torr) x (28.01344 g N2/mol) = 0.433 g N2 
(0.033769 mol) x (131 torr O2/ 483 torr) x (31.99886 g O2/mol) = 0.293 g O2 
(0.033769 mol) x (131 torr He/ 483 torr) x (4.0026 g He/mol) = 0.0367 g He

Answer;

  • 0.433 g N2  
  • 0.293 g O2
  • 0.0367 g He

Explanation and solution;

  • We can start by getting the total pressure; which will be the sum of the partial pressure of each gas.

221 torr + 131 torr + 131 torr

P (total)  = 483 torr total  

  • Using the equation

n = PV / RT, we can determine the total number of moles of the mixture

= (483 torr) x (1.30 L) /  ((62.36367 L Torr/K mol) x (25.0 + 273.15 K))

= 0.033769 mol gases total  

  • Therefore; we can determine the mass of each gas;
  • Nitrogen gas

N2 = 28.01 g/mol

= (0.033769 mol) x ( 221 torr N2/ 483 torr) x (28.01 g N2/mol)

=  0.433 g N2  

  • Oxygen gas

O2 = 32  g/ mol

=(0.033769 mol) x (131 torr O2/ 483 torr) x (32 g O2/mol)

= 0.293 g O2  

  • Helium gas

He = 4 g/mol

= (0.033769 mol) x (131 torr He/ 483 torr) x (4.00 g He/mol)

= 0.0367 g He

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