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

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

How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL Cl2(g) at 25 °C and 745 Torr ?

mass of MnO2:

Answer 1

0.605 g

Step-by-step explanation:

• MnO₂(s) + 4HCl(aq) ⟶ MnCl₂(aq) + 2H₂O(l) + Cl₂(g)

First we calculate how many Cl₂ moles need to be produced, using the PV=nRT formula:

• P = 745 Torr ⇒ 745 / 760 = 0.980 atm

• V = 185 mL ⇒ 185 / 1000 = 0.185 L

• n = ?

• R = 0.082 atm·L·mol⁻¹·K⁻¹

• T = 25 °C ⇒ 25 + 273.16 = 298.16 K

Inputting the data:

• 0.980 atm * 0.185 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

• n = 0.00696 mol

Then we convert 0.00696 moles of Cl₂ to MnO₂ moles:

• 0.00696 mol Cl₂ *
  `(1molMnO_2)/(1molCl_2)` = 0.00696 mol MnO₂

Finally we convert 0.00696 moles of MnO₂ to grams, using its molar mass:

• 0.00696 mol MnO₂ * 86.94 g/mol = 0.605 g