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 225 mL Cl2(g) at 25 °C and 785 Torr?

mass:___________g MnO2

Answer 1

Given:

P = 785.0 torr

= (785.0/760) atm

= 1.0329 atm

V = 225.0 mL

= (225.0/1000) L

= 0.225 L

T = 25.0 oC

= (25.0+273) K

= 298 K

find number of moles using:

P * V = n*R*T

1.0329 atm * 0.225 L = n * 0.08206 atm.L/mol.K * 298 K

n = 9.504*10^-3 mol

This is mol of Cl2 formed

From reaction,

mol of MnO2 reacted = mol of Cl2 formed

= 9.504*10^-3 mol

Molar mass of MnO2,

MM = 1*MM(Mn) + 2*MM(O)

= 1*54.94 + 2*16.0

= 86.94 g/mol

use:

mass of MnO2,

m = number of mol * molar mass

= 9.504*10^-3 mol * 86.94 g/mol

= 0.8263 g

Answer: 0.826 g