Question

5.80 mol of solid A was placed in a sealed 1.00-L container and allowed to decompose...

5.80 mol 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 1.20 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?​

3.40 mol

Explanation

A(s) <-------> B(g) + C(g)

Kc = [B][C]

at equillibrium ,

[B] = x

[C] = x

at equillibrium , [B] = [C] = 1.20M

Kc = 1.20 × 1.20 = 1.44

After changing the volume

Initial concentration

[B]= 1.20M/2 = 0.60M

[C] = 1.20M /2 = 0.60M

Change in concentration

[B] = +x

[C] = +x

Equillibrium concentration

[B] = 0.60 + x

[C] = 0.60 +x

so,

(0.60 + x)2 = 1.44

0.60 + x = 1.2

x = 0.60

[ B] = 0.60 + 0.60 = 1.20M

[C] = 0.60 + 0.60 = 1.20M

number of moles of B = (1.20mol/1.00L ) × 2.00L = 2.4 moles

number of moles of C = (1.20mol/1.00L) × 2.00L = 2.4moles

2.4moles of B and 2.4moles of C are comes from the decomposition of 2.4 moles of A

Therfore

remaining moles of A = 5.80mol - 2.40mol= 3.40mol

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