5.00 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 1.10 M, where it remained constant. Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain?
A(s)--> B(g)+C(g)
just multiply the concentration of
Bx2=1.10Mx2L=2.2moles
and then subtract the moles of A from calculated moles of B
5.00-2.2=2.8moles
To find out how many moles of A remain, we need to use the given information about the concentration of B and the change in container volume.
We know that the initial concentration of B, before the volume change, was 1.10 M and it remained constant even after the volume change. Let's denote the concentration of B before the volume change as [B]_1.
Now, for a reaction in a closed container, the product concentration is related to the reactant concentration by the stoichiometric coefficients of the balanced equation. In this case, the stoichiometric coefficient of B is 1, which means that the concentration of B is directly proportional to the concentration of A.
Since the concentration of B remained constant, we can conclude that the concentration of A also remained constant after the volume change. Let's denote the concentration of A after the volume change as [A]_2.
Since the volume of the container doubled, the concentration of A will also double since the amount of A remains constant. Therefore, we can write:
[A]_2 = 2[A]_1
Now, the number of moles is directly proportional to the concentration, and since the volume doubled, the concentration of A also doubled. So, the number of moles of A before the volume change is equal to the number of moles of A after the volume change.
Let's denote the number of moles of A before the volume change as n_A1 and the number of moles of A after the volume change as n_A2.
n_A1 = n_A2
Now, we can set up a proportion between the concentrations:
[A]_1 / [A]_2 = n_A1 / n_A2
Substituting the values we know, we have:
1.10 M / 2.20 M = 5.00 mol / n_A2
Simplifying the equation, we find:
n_A2 = 2.27 mol
Therefore, 2.27 moles of A remain in the container.