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 385 mL Cl 2 ( g ) at 25 °C and 795 Torr ?

929 mg of MnO₂ should be added to excess HCl, in order to obtain 385 mL of Cl₂

Explanation:

The reaction states:

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

First of all we need the moles of produced chlorine. We can find it out by the Ideal Gases Law → P . V = n . R . T

- We need to convert the 795 Torr to atm →  795 Torr . 1 atm / 760 Torr = 1.05 atm

- We need to convert the 385 mL to L → 385 mL . 1L / 1000mL = 0.385L

- We need T° K = 25°C + 273 = 298K

1.05 atm . 0.385L = n . 0.082 . 298K

n = 1.05 atm . 0.385L / 0.082 . 298K → 0.0165 moles of Cl₂  

Ratio is 1:1. 1 mol of chlorine gas can be made by 1 mol of MnO₂

Therefore, If I produced 0.0165 moles of Cl₂  I definitely used 0.0165 moles of MnO₂

We convert the moles to mass → 0.0165 mol . 86.94 / 1 mol = 1.43 g

We can convert the g to mg → 1.43 g . 1000 mg / 1g = 1430 mg

We need 1.43 grams of MnO2

Explanation:

Step 1: Data given

Volume of Cl2 = 385 mL

Temperature = 25 °C = 298 K

Pressure = 795 torr = 1.04605 atm

molar mass of MnO2 = 86.93 g /mol

Step 2: The balanced equation

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

Step 3: Calculate moles Cl2

p*V = n*R*T

n = (p*V)/(R*T)

⇒n = number of moles Cl2 = TO BE DETERMINED

⇒p = the pressure of Cl2 = 1.04605 atm

⇒V = the volume of Cl2 = 0.385 L

⇒R = the gas constant = 0.08206 L*atm/mol*K

⇒T = the temperature = 25 °C = 298 K

n = (1.04605 * 0.385)/(0.08206 * 298)

n = 0.0165 moles

Step 4: Calculate moles MnO2

For 1 mol MnCl2, 2 moles H2O and 1 mol Cl2 produced, we need 1 mol MnO2 and 4 moles HCl

For 0.0165 moles Cl2 we need 0.0165 moles MnO2

Step 5: Calculate mass of MnO2

Mass MnO2 = moles MnO2 * molar mass MnO2

Mass MnO2 = 0.0165 moles * 86.93 g/mol

Mass MnO2 = 1.43 grams

We need 1.43 grams of MnO2

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