Posted by **Sandra** on Friday, May 11, 2012 at 1:32am.

A 100 mL container is filled with 0.01 mol of A2B3. At equilibrium, it also contains 0.006 mol of B. Find the value of K.

A2B3(g) --> 2A(g) +3B(g)

- Chemistry -
**DrBob222**, Friday, May 11, 2012 at 3:51pm
........A2B3 ==> 2A + 3B

I......0.01.......0....0

C.......-x........2x...3x

E....0.01-x.......2x...3x

The problem tells us that 3x = 0.006 = mols B; therefore, x = 0.006/3 = 0.002

Equilibrium mols are as follows:

A2B3 = 0.01-0.002 = ?

A = 2*0.002 = ?

B = 0.006

These are mols. You must convert to molarity. Do that by M = mols/L. The volume is given as 0.1L. Then substitute concns into the Keq expression and solve for Kc.

## Answer This Question

## Related Questions

- Chemistry - PLEASE fully explain these...I've already tried to figure them out ...
- Chemistry - A2B3 --> 2A + 3B I 0.001 0 0 C –x 2x 3x E 0.001-x 2x 3x 3x = 0....
- AP Chemistry - The equilibrium constant for thermal dissociation of F2 F2(g)<...
- chemistry - In a 1.00 L container at 272 degrees C 1.3 mol of N2 and 1.65 mol of...
- chem 12 - 1.0 mol of each of the gases, CO, H20, CO2 and H2 are placed in a 2....
- chem12 - 1.0 mol of each of the gases, CO, H20, CO2 and H2 are placed in a 2.00L...
- chemistry - 1.00 mol of A and 2.00 mol of B are placed in a 5 Liter container. ...
- chem 12 - consider the following equilibrium 2NOCl(g)--- 2 NO(g) + Cl2(g) ...
- chem - A solution initially contains 0.350 mol/L of A and 0.750 mol/L of B. A ...
- Chemistry - An equilibrium mixture in a 10.0 L flask contains 7.0 mol HI(g) and ...

More Related Questions