A 5.20 mol sample of solid A was placed in a sealed 1.00 L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.30 M, where it remained constant. A(s)↽−−⇀B(g)+C(g) A ( s ) ↽ − − ⇀ B ( g ) + C ( g ) Then, the container volume was doubled and equilibrium was re‑established. How many moles of A A remain?

2.60 moles of A remaining.

Explanation:

According to  Le Chatelier's principle, the equilibrium would shift if the volume, concentration, pressure, or temperature changes.

In this question, we were told that  the volume doubles, that implies that we would have to double the molarity of B/ C (since B=C.)

However, it is obvious and clear from the given equation of the reaction that  A is  solid in it's activity = 1. Hence, it is then ignored.

So doubling B would be 1.30 M × 2 = 2.60 M

i.e 2.60 M moles of A was consumed.

Now;  the number of moles of A remaining  is 5.20 - 2.60 = 2.60 moles of A remaining.

i believe the answer is true.