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
