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) +4HCI(aq) → MnCl₂(aq)+2H₂O(I) + Cl₂(g)

How much MnO₂(s) should be added to excess HCI(aq) to obtain 405 mL of Cl₂(g) at 25 °C and 705 Torr?

Answer 1

To calculate the amount of MnO2(s) needed to produce 405 mL of Cl2(g) at 25 °C and 705 Torr, you can use the ideal gas law and stoichiometry.

Step-by-step explanation:

To calculate the amount of MnO2(s) needed to produce 405 mL of Cl2(g) at 25 °C and 705 Torr, we can use the ideal gas law. First, we need to convert the volume of Cl2(g) to moles using the ideal gas law equation:

n = PV / RT

Where:

• n is the number of moles of Cl2(g)
• P is the pressure of Cl2(g) in atm
• V is the volume of Cl2(g) in liters
• R is the ideal gas constant (0.0821 L.atm/mol.K)
• T is the temperature of Cl2(g) in Kelvin

Once we have the number of moles of Cl2(g), we can use the stoichiometry of the balanced chemical equation to determine the moles of MnO2(s) needed. From the balanced equation, we know that 1 mole of MnO2(s) produces 1 mole of Cl2(g). Therefore, the amount of MnO2(s) needed is equal to the number of moles of Cl2(g).