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Suppose a mixture containing 3.93 g of [tex] \text{H}_2 [/tex] and 38.4 g of [tex] \text{NO} [/tex] has a total pressure of 3.01 atm. What are the partial pressures of both gases in the mixture?

Answer :

Answer and explanation

We are asked to calculate the partial pressure of each gas given the mass of the gasses

We can do this as follows

For H2

Mols = 3.93g/2.016 g/mol

= 1.95 mols

For NO

Mols = 38.4 g/ 30.01 g/mol

= 8.31 mols

the total number of mols then becomes:

8.31 + 1.95 =

10.26

and the mol fraction of each gas is determined by dividing the number of mols of that gas by the total number of mols

for H2 = 1.95/10.26

= 0.190

for NO = 8.31/10.26

= 0.809

now that we have the mols we can determine the pressures exerted by each gas by multiplying the total pressure by the mol fraction of each gas

for H2 the partial pressure is:

P(H2) = 3.01 atm x 0.190

= 0.572 atm

for NO the partial pressure is:

P(NO) = 3.01 atm x 0.809

= 2.44 atm

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