Question

A 4.80 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 moles of A remains are, 2.00 mole

Explanation :

As we are given that,

Number of moles of solid A = 4.80 mole

Volume of container = 1.00 L

Concentration of A =
`(4.80mole)/(1.00L)=4.80mole/L`

Concentration of B = 1.40 M

The moles of B in 1 liter of solution =
`1.40mole/L* 1L=1.40mole`

When the volume is doubled then we are using volume 2 L instead of 1 L.

The moles of B in 2 liter of solution =
`1.40mole/L* 2L=2.80mole`

The balanced chemical reaction is,

`A(s)\rightleftharpoons B(g)+C(g)`

initial conc. 4.80 0 0

At eqm. (4.80-2.80) 2.80

The moles of A remains = 4.80 - 2.80 = 2.00 mole

Therefore, the number of moles of A remains are, 2.00 mole