the reaction of MnO2(s)+4HCl(aq)--- MnCl2(aq)+Cl2(g)+H2O(l) was being studied.12.2g of MnO2 were reacted with 70 ml of 2.5 m HCl solution, the chlorine prod0.971 atm at temp of 78.2C

Answer 1

balance equ:

MnO2(s)+4HCl(aq)⇐⇒MnCl2+Cl2+2H2O

1 mole MnO2 reacts with 4 HCl and produces 1 mole Cl2.

⇒12.2 grams of Mno2= (12.2/87) mol = 0.140 mol MnO2;

⇒ 70 ml 2.50M HCl = (70/1000)∗2.5 mol = 0.175 mol HCl

Since Mno2 is excess, 0.175 mol HCl is totally consumed by (0.175/4)= 0.04375 mol MnO2 to liberate (0.175/4)= 0.04375 mole Cl2

⇒MnO2 left = (0.140 - (0.175/4))moles

⇒Moles of Cl2 = (0.175/4)= 0.04375

We know,

PV = nRT

V = (nRT/P)

V= ((0.175/4)∗(0.0821 L.atm/k)∗(78.2+273))/(0.971 atm)

⇒V = 1.297 L