# N_{2} + 3H_{2}\rightarrow2NH_{3} After completion of 75% of the reaction what would be the mole fraction for the constituents

As we learnt in

Mole Fraction -

It is the ratio of moles of solute or moles of solvent to moles of the solution.

- wherein

If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are $n_{A}$ and $n_{B}$ respectively; then the mole fractions of A and B are given as

Mole fraction of A = (number of moles of A)/(number of moles of solution ) = $n_{A}$/($n_{A}$+ $n_{B}$)

At   $t = \frac{3}{4}$,

$N_{2} + 3H_{2}\rightarrow2NH_{3}$

$t = \frac{3}{4}$               $\frac{1}{4}$            $\frac{3}{4}$                $\frac{3}{2}$

Moles of N2 = 0.25

Moles of H2 = 0.75

Moles of NH3 = 1.5

Mole fraction of  $N_{2}=\frac{0.25}{2.5} =0.1$

Mole fraction of  $H_{2}=\frac{0.75}{2.5} =0.3$

Mole fraction of  $NH_{3}=\frac{1.5}{2.5} =0.6$

Exams
Articles
Questions