Respuesta :
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
The number of moles of A that remain is; 2.8 moles
We are given the decomposition reaction as;
A(s) ⇆ B(g) + C(g)
Also;
Number of moles of solid A; n_a = 5.6 mol
Volume of the container; V = 1 L
Concentration of gas B = 1.4 M
Formula for concentration is;
Concentration = number of Moles/Volume
Thus, concentration of solid A is;
Cₐ = 5.6 moles/1 L
Cₐ = 5.6 M
Similarly;
No of moles = Concentration × volume
Thus, number of moles of gas B is;
n_b = 1.4 × 1
n_b = 1.4 mol
We are told that the volume gets doubled. Thus, V' = 2 L
n'b = 1.4 × 2
n'b = 2.8 moles
Now, the Initial concentration of B and C is 0 M.
Thus;
Cₐ at equilibrium = 5.60 - 2.80
Cₐ at equilibrium = 2.8 M
Thus, number of moles of A at equilibrium is 2.8 moles
∴ Remaining moles of A = 5.6 moles - 2.8 moles
Remaining moles of A = 2.8 moles
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