For the reaction [tex]A(g) + B(g) \longrightarrow AB(g)[/tex] , there are 21412 unique collisions between A and B are possible if 1.01 mol of A(g) and 2.12 mol of B(g) are present in the vessel .
The given reaction is, A(g) + B(g) → AB(g)
In this question it has been given that, 1.01 mol of A and 2.12 mol of B present in the vessel.
Hence we can write the balance equation as,
1.01 A(g) + 2.12B(g) → AB(g)
Or 101/100 A(g) + 212/100B(g) → AB(g)
101 A(g) + 212B(g) → 100AB(g)
Here each A particle can collide with 212 mol of B particles present in the vessel. So the possible unique collisions are, 101×212 = 21412
Thus 21412 unique collisions between A and B are possible if there are 1.01 mol of A and 2.12 mol of B present in the vessel.
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