chemistry homework

Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described by the chemical equation

MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

How much MnO2(s) should be added to excess HCl(aq) to obtain 165 mL Cl2(g) at 25 °C and 795 Torr ?

Answer 1

MnO₂(s) + 4HCl(aq)⟶MnCl₂(aq) + 2H₂O(l) + Cl₂(g) ... balanced equation

Because the HCl is present in excess, we need only concern ourselves with the moles of MnO₂ 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 Cl₂ is derived from 1 mole of MnO₂.

To answer this question, we will first determine how many moles of Cl₂ are present in 165 ml of Cl₂ 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 Cl₂

To make 0.00705 mol Cl₂ will require 0.00705 mol MnO₂

Mass MnO₂ needed = 0.00705 mol MnO₂ x 86.9 g/mol = 0.613 g MnO₂