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2.0 g of N2, H2 and 0.50 g of 9.0 g of O2 are placed in...

2.0 g of N2, H2 and 0.50 g of 9.0 g of O2 are placed in a container of 1.00 L to 27 degrees C, determine the total pressure in the container (presumed ideal gas PV = nRT). using the law of Dalton calculate the partial pressure of O2.

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Answer #1

given are the weights of nitrogen, hydrogen and oxygen gases.

we can calculate the number of moles of each gas as

number of moles = weight/molecular weight

number of moles of N2 (n1) = 2/28 = 0.0714 mol

number of moles of H2 (n2)= 0.5/2 = 0.25 mol

number of moles of O2 (n3)= 9/32 = 0.28125 mol

hence total number of moles(n) = 0.0714 + 0.25 + 0.28125 = 0.6 mol

now use ideal gas equation,

PV = nRT

P1 = 0.60.0821300 (temperature T = 27+273 = 300K)

P = 14.778 atm

now for partial pressure of O2

use dalton law,

Pi = yiP (here yi is mole fraction)

P3 = (0.28125/0.6)14.778

P3 = 6.9 atm

hence partial pressure of O2 is 6.9 atm

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