A chemical reaction occurring in a cylinder equipped with a movable piston produces 0.58 mol of a gaseous product. If the cylinder contained 0.11 mol of gas before the reaction and had an initial volume of 2.1 L, what was it's volume after the reaction.

Am I doing this right
V1/n1=V2/n2
I'm looking for V2
0.58 mol/0.11mol. 2.1L = 11.0L + 2.1L = 13L did I do this right

Does the 0.11 mol there initially enter into the reaction? If not, then you are right at about 13L.

Yes, you are on the right track! The equation you're using, V1/n1 = V2/n2, is derived from the ideal gas law equation (PV = nRT), where V is the volume, n is the number of moles, P is the pressure, R is the ideal gas constant, and T is the temperature. So, in this case, you can use the ratio of moles and volumes to find the final volume.

To calculate V2, you correctly set up the equation V1/n1 = V2/n2. However, there seems to be a slight mistake in your calculation. Let's correct it step by step:

Given:
V1 = 2.1 L (initial volume)
n1 = 0.11 mol (initial moles)
n2 = 0.58 mol (final moles)

Plug in the values:
V1/n1 = V2/n2
2.1 L / 0.11 mol = V2 / 0.58 mol

Next, you can solve for V2 by cross-multiplication:
(2.1 L) * (0.58 mol) = (0.11 mol) * V2

Multiply and divide:
V2 = (2.1 L * 0.58 mol) / 0.11 mol

Now, calculate the expression:
V2 = 2.1 L * 0.58 / 0.11
V2 = 11.4 L

Therefore, the volume after the reaction is approximately 11.4 L.