Question

A 5.60 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 reaches 1.40 M, where it remained constant. A(s)↽−−⇀B(g)+C(g) Then, the container volume was doubled and equilibrium was re‑established. How many moles of A remain?

Answer 1

Answer:

The number of remaining moles of A = 2.80 moles

Explanation:

Step 1: Data given

A(s) ⇆ B(g)+C(g)

Number of moles of solid A = 5.60 mol

Volume of the container = 1.00 L

The concentration of B steadily increased until it reaches 1.40 M

The container volume was doubled and equilibrium was re‑established.

Step 2: Calculate concentration of solid A

Concentration = Moles / Volume

Concentration = 5.6 moles / 1.00 L

Concentration = 5.6 M

Step 3: Calculate moles of B

Moles = Concentration * volume

Moles = 1.40 M *1.00 L = 1.40 moles

Step 4: The volume gets doubled

Moles B = 1.40M * 2.00 L = 2.80 moles

Step 5: The balanced equation

A(s) ⇆ B(g)+C(g)

Initial concentration of A = 5.6M

Initial concentration of B and C = 0 M

Concentration of A at the equilibrium = 5.60 - 2.80 = 2.80

Remaining moles of A = 2.80 moles

The number of remaining moles of A = 2.80 moles