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
MnO 2 ( s ) + 4 HCl ( aq ) ⟶ MnCl 2 ( aq ) + 2 H 2 O ( l ) + Cl 2 ( g )

How much MnO 2 ( s ) should be added to excess HCl ( aq ) to obtain 255 mL Cl 2 ( g ) at 25 °C and 795 Torr ?

Answer 1

Answer:

0.9483 grams of manganese dioxide should be added to excess HCl.

Explanation:

Pressure of the chlorine gas = P = 795 Torr = 1.046 atm  (1 atm = 760 Torr)

Volume of the chlorine gas = V = 255 ml = 0.255 L

Temperature of the chlorine  gas = T = 25°C= 298.15 K

Moles of chlorine gas = n

Using ideal gas equation:

PV = nRT

$n=\frac{PV}{RT}=\frac{1.046 atm\times 0.255 L}{0.0821 atm l/mol k\times 298.15 K}$

n = 0.01090 mol

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

According to reaction , 1 mole of chlorine gas is obtained from 1 mole of manganese dioxide.

Then 0.01090 moles of chlorine gas will be obtained from:

$\frac{1}{1}\times 0.01090 mol=0.01090 mol$ manganese dioxide

Mass of 0.01090 moles of manganese dioxide:

0.01090 mol × 87 g/mol = 0.9483 g