Chlorine can be set up in the research facility by the response of manganese dioxide with hydrochloric corrosive, HCl(aq), as portrayed by the concoction condition

MnO2(s)+4HCL(aq)- - >MnCl2(aq)+ 2H2O (l)+ Cl2 (g)

The amount MnO2(s) ought to be added to overabundance HCl(aq) to get 115 mL of Cl2(g) at 25 °C and 785.

What volume would 115 cm3 of Cl2 (g) at 25 °C (298 K) and 785 torr have at Standard Temperature and Pressure, STP, (273 K and 760 torr)?

P1V1/T1 = P2V2/T2

Adjust:

V2 = P1V1T2/P2T1

Where:

V1 = 115 cm3

V2 = volume at STP

P1 = 785 torr

P2 = 760 torr

T1 = 298 K

T2 = 273 K

So: V2 = (785 torr x 115 cm3 x 273 K)/(760 torr x 298 K) = 108.8 cm3

Presently at STP 1 mole of a gas has a volume of 22.4 dm3.

So: 108.8 cm3 (0.1088 dm3) of Cl2 (g) will contain 0.1088 dm3/22.4 dm3/mol = 0.005 mol

From the condition 1 mol of Cl2 (g) is delivered by 1 mol of MnO2 (s)

So: 0.005 mol of Cl2 (g) is delivered by 0.005 mol of MnO2 (s)

Relative nuclear masses:

Mn = 55

O = 16

At that point 1 mol of MnO2 = 55 g + (16 g x 2) = 87 g/mol

So the MnO2 (s) required = 0.005 mol = 87 g/mol x 0.005 mol = 0.435 g