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 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 ?

Answer 1

Answer:

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 = $765 Torr=\frac{765}{760}atm$

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:

$(\frac{765 }{760}atm)\times 0.385 L=n\times (0.0821L.atm/mol.K)\times 300K\\\\n=0.01573 mole$

$MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)$

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 :

$\frac{1}{1}\times 0.01573 mol=0.01573 mol$ 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.