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

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

Step-by-step 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=(PV)/(RT)=(1.046 atm* 0.255 L)/(0.0821 atm l/mol k* 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:

`(1)/(1)* 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