Chemistry, molar volume problem, read details please

MnO2 + 4 HCl --> MnCl2 + CL2 + 2 H2O

a) The volume of CL2 produced from the reaction of 10 grams of HCL with excess MnO2

b) The volume of CL2 from the reaction of 20 ml of HCL with excess MnO2

Without more information, one of those parts isn't doable. If you are talking about HCl in its gas form,then a is doable. If in a liquid form then b is not. In fact, no matter what the form, B is not possible.

A]
Balance the equation (which is done) Assume that you are working with gaseous HCl

1 mol of HCl has a mass of 1 + 35.5 grams = 36.5
x mol has a mass of 10 grams

36.5 x = 10 grams
x =.274 mol

Next find the number of mols of Cl2
4 mol HCl produces 1 mol Cl2
0.274 mol produces x mol Cl2

4/0.274 = 1/x
4x = 0.274
x = 0.0685

At STP 1 mol of any gas occupies 22.4 L
0.0685 mol occupies x
1/0.0685 = 22.4 / x
x = 0.0685 * 22.4
x = 1.5344 L