the reaction of 4A+3B→2C+D is studied. Unknown masses of the reactants were mixed . After a reaction time of 1 hour the analysis of the mixture showed 2 kmol, 1 kmol of B and 4 kmol of C. product D was present in the mixture but could not be analysed. what is the mole fraction of D in the mixture?

the mole fraction of D in the mixture is 0.2222

Explanation:

Given that;

mixture analysis shows 2 kmol A, 1 kmol B, 4 kmol C and some unknown kmol of D was present.

4A+3B→2C+D

As from reaction stoichiometry, for every 2 kmol of C produced, kmol of D produced = 1 kmol

so, for 4 kmol C, kmol of D produced = 4/2 × 1 kmol = 2 kmol

Now our mixture has 2 kmol A, 1 kmol B, 4 kmol C and also 2 kmol of D

so, total moles in mixture, we have (2 + 1 + 4 + 2) kmol = 9 kmol

mole fraction of D in mixture  will be;

( Kmol of D) / (total moles in mixture) = 2 / 9 = 0.2222

Therefore the mole fraction of D in the mixture is 0.2222

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