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,(s) + 4 HCl(aq)
MnCl(aq) + 2 H2O(l) + Cl (8)
How much MnO(s) should be added to excess HCl(aq) to obtain 175 mL C12(g) at 25 °C and 715 Torr?
mass of MnO2
​

Answer 1

Step-by-step explanation:

MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g 22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

= 175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .