Question

One mole of a mixture of n2 , no2 and n2o4 has a mean molar mass of 55.4. on heating to a temperature at which all the n2o4 may be presumed to have dissociated : n2o4 2no2 , the mean molar mass tends to the lower value of 39.6. what is the mole ratio of n2 : no2 : n2o4 in the original mixture ?

Answer 1

Let the number of moles of N₂ be X, the number of moles of NO₂ be y and number of moles of N₂O₄ be Z.

Total moles = 1

So x +y +z = 1 Equation 1

Molar mass of N₂ = 28

Molar mass of NO₂= 46

Molar mass of N₂O₄ = 92

Mean average of the molar mass = $\frac{28x + 46 y + 92z}{x +y +z}$ = 55.6

Using the relation x + y +z =1,

28x + 46 y + 92z = 55.6 Equation 2

moles of NO₂ formed from N₂O₄ = 2z

Hence after reaction, NO₂ = y + 2z

Now mean average of the molar mass after reaction = $\frac{28x + 46 y + 92z}{x + y +2z}$ = 39.6 Equation 3

Solving the three equations

We get

x = 0.5

y = 0.1

z = 0.4

Thus the ratio becomes x : y : z = 0.5 : 0.1 : 0.4

Answer 2

Answer:- x : y : z = 0.5 : 0.1 : 0.4

Explanation:

Let the number of moles of N₂ be X, the number of moles of NO₂ be y and number of moles of N₂O₄ be Z.

Total moles = 1

So x +y +z = 1 Equation 1

Molar mass of N₂ = 28

Molar mass of NO₂= 46

Molar mass of N₂O₄ = 92

Mean average of the molar mass = = 55.6

Using the relation x + y +z =1,

28x + 46 y + 92z = 55.6 Equation 2

moles of NO₂ formed from N₂O₄ = 2z

Hence after reaction, NO₂ = y + 2z

Now mean average of the molar mass after reaction = = 39.6 Equation 3

Solving the three equations

We get

x = 0.5

y = 0.1

z = 0.4

Thus the ratio becomes x : y : z = 0.5 : 0.1 : 0.4

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