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 MnO 2 ( s ) + 4 HCl ( aq ) ⟶ MnCl 2 ( aq ) + 2 H 2 O ( l ) + Cl 2 ( g ) MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+ Cl2(g) How much MnO 2 ( s ) MnO2(s) should be added to excess HCl ( aq ) HCl(aq) to obtain 385 mL Cl 2 ( g ) 385 mL Cl2(g) at 25 °C and 765 Torr 765 Torr ?

1.368 grams of manganese dioxide must b added to HCl to obtain 385 mL of chlorine gas.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of chlorine gas =

1 atm = 760 Torr

V = Volume of chlorine gas = 385 mL = 0.385 L ( 1 mL - 0.001 L)

n = number of moles of chlorine gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of chlorine gas =25 °C= 25 + 273 K =  300 K

Putting values in above equation, we get:

According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese dioxide ,then 0.01573 moles of chlorine gas will be obtained from :

of manganese dioxide

Mass of 0.01573 moles of manganese dioxide:

0.01573 mol × 86.94 g/mol = 1.368 g

1.368 grams of manganese dioxide must b added to HCl to obtain 385 mL of chlorine gas.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.