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 ?

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 = 0.01090 mol

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:

manganese dioxide

Mass of 0.01090 moles of manganese dioxide:

0.01090 mol × 87 g/mol = 0.9483 g

i think   the answer is c hope i'm right