J.R. S. answered • 11/26/20

Ph.D. University Professor with 10+ years Tutoring Experience

MnO_{2}(s) + 4HCl(aq)⟶MnCl_{2}(aq) + 2H_{2}O(l) + Cl_{2}(g) ... balanced equation

Because the HCl is present in excess, we need only concern ourselves with the moles of MnO_{2} that are present. This will determine the theoretical yield of Cl2. Note that they are in a mole ratio of 1:1, i.e. 1 mol Cl_{2} is derived from 1 mole of MnO_{2}.

To answer this question, we will first determine how many moles of Cl_{2} are present in 165 ml of Cl_{2} at 25ºC and 795 torr pressure. To do this, we'll use the ideal gas law, PV = nRT and solve for n (moles):

n = PV/RT where

P = pressure in atm = 795 torr x 1 atm/760 torr = 1.046 atm

V = volume in L = 165 ml x 1 L/1000 ml = 0.165 L

R = gas constant = 0.0821 Latm/Kmol

T = temp in Kelvin = 25ºC + 273 = 298K

n = (1.046)(0.165) / (0.0821)(298) = 0.00705 moles Cl_{2}

To make 0.00705 mol Cl_{2} will require 0.00705 mol MnO_{2}

Mass MnO_{2} needed = 0.00705 mol MnO_{2} x 86.9 g/mol = **0.613 g MnO**_{2}