How many liters of chlorine gas at 25C and 0.950 atm can be produced by the reaction of 24.0 g of MnO2 with hydrochloric acid? The products include manganese (II) Chloride, water, and chlorine.

Answer 1

You have posted in the last day or so several problems that are similar. They are stoichiometry problems and all of those are worked the same way i a 4-step process. Here are the steps with this problem. I suggest you print and save this.
1. Write and balance the equation.
MnO2 + 4HCl ==> MnCl2 + Cl2 + 2H2O

2. Convert what you have (in this case 24.0 g MnO2) to mols. mol = grams/molar mass = ?

3. Using the coefficients in the balanced equation, convert mols of what you have (MnO2) to mols of what you want (in this case mols of what you want is mols Cl2). You do it this way.
mols Cl2 = mols MnO2 x (1 mol Cl2/1 mol MnO2) = mols MnO2 x 1 = ?; mols Cl2 = mols MnO2.

4. Now convert mols Cl2 to whatever unit you wish.
a. Most problems ask for grams Cl2. That is grams = mols x molar mass = ?
b. This problem asks for volume. So use PV = nRT, substitute P and T from the problem and solve for volume in liters. Don't forget T must be in kelvin.

Post your work if you get stuck.

Answer 2

Thanks. I see what I did wrong now, the T is 298 K not 273 K.