Question

Cl2 gas is prepared by the reaction of Manganese dioxide with hydrochloric acid according to the following equation. Calculate the number of moles of chlorine gas produced by the reaction of 43.5 g MnO2 and 9 g of HCl. Also calculate the volume of Cl2 produced at STP.

MnO2 + 4HCl —; MnCl2 + 2H2O + Cl2

Answer 1

Explanation:

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