Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq) , as described by the chemical equation MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) How much MnO2(s)MnO2(s) should be added to excess HCl(aq)HCl(aq) to obtain 245 mL Cl2(g)245 mL Cl2(g) at 25 °C and 705 Torr705 Torr ?

Answer 1

Answer:

0.81 grams MnO2

Explanation:

Solve this problem by using the stoichiometry of the balanced chemical reaction.

We are given the volume of Cl2 produced  which can be converted to the number of moles of Cl by utilizing the ideal gas law and then since the balanced reaction show us that 1 mol of chlorine will be produced from 1 mol MnO2 it follows we will know the moles of MnO2 and then multiply it by its molecular weight to answer the question.

PV= nRT  then n= PV/RT

P: 705 torr x 1 atm760 torr = 0.93 atm

V: 245 mL x  1 L/1000 mL = 0.245 L

R= 0.08205 LAtm/Kmol

T: 25 ºC + 273 = 298 K

(note the values of V, P and T needed to be converted according to the units in the R constan)

n= 0.93 atm x 0.245 L/( ( 0.08205 Latm/Kmol) x 298 K)) = 0.009 mol Cl2

0.009 mol Cl2 x 1 mol MnO2/mol Cl2 = 0.009 mol MnO2

0.009 mol MnO2 x 86.93 g/mol MnO2 = 0.81 g MnO2