What volume of chlorine can be obtained by the interaction of 2 mol of hydrogen chloride and 3 mol

What volume of chlorine can be obtained by the interaction of 2 mol of hydrogen chloride and 3 mol of manganese oxide 4?

Given:
n (HCl) = 2 mol
n (MnO2) = 3 mol

To find:
V (Cl2) -?

1) Write the equation of a chemical reaction:
4HCl + MnO2 => Cl2 ↑ + MnCl2 + 2H2O;
2) Compare the amount of HCl and MnO2 substance:
The amount of HCl is in short supply, therefore, calculations will be carried out for it;
3) Determine the amount of substance Cl2 (taking into account the coefficients in the reaction equation):
n (Cl2) = n (HCl) / 4 = 2/4 = 0.5 mol;
4) Calculate the volume of Cl2 (under normal conditions):
V (Cl2) = n (Cl2) * Vm = 0.5 * 22.4 = 11.2 liters.

Answer: The volume of Cl2 is 11.2 liters.