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)

Question

How much MnO2(s) should be added to excess HCl(aq) to obtain 285 mL Cl2(g) at 25 °C and 705 Torr?

Answer 1

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

We'll first find how many moles of Cl2 gas is to be obtained:

PV = nRT

P = pressure = 705 torr x 1 atm/760 torr = 0.928 atm

V = volume = 285 ml = 0.285 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 25C + 273 = 298K

n = moles = PV/RT = (0.928)(0.285) / (0.0821)(298)

n = 0.0108 moles of Cl₂

Now, using dimensional analysis and the stoichiometry of the balanced equation, we solve for moles and grams of MnO₂ needed...

0.0108 mol Cl₂ x 1 mol MnO₂ / mol Cl₂ = 0.0108 moles MnO₂ needed

0.0108 mol MnO₂ x 86.9 g/mol MnO₂ = 0.939 g MnO₂ needed