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 HCl(aq) to obtain 365 mL Cl₂(g) at 25°C and 725 Torr?

Answer 1

To calculate the amount of MnO₂(s) needed to obtain 365 mL of Cl₂(g), we can use the stoichiometry of the reaction. Convert the volume of Cl₂(g) to moles using the ideal gas law, find the moles of MnO₂(s) using the stoichiometric ratio, and then convert moles of MnO₂(s) to grams using the molar mass.

Step-by-step explanation:

To calculate the amount of MnO₂(s) required to obtain 365 mL of Cl₂(g) at 25°C and 725 Torr, we can use the stoichiometry of the reaction. First, convert the volume of Cl₂(g) to moles using the ideal gas law. Then use the stoichiometric ratio between MnO₂(s) and Cl₂(g) to find the moles of MnO₂(s). Finally, convert moles of MnO₂(s) to grams using the molar mass of MnO₂(s).

Let's assume that the reaction is conducted at standard temperature and pressure (STP) and the molar volume of a gas is 22.4 L/mol.

Given:
- Volume of Cl₂(g) = 365 mL = 0.365 L
- Temperature = 25°C = 298 K
- Pressure = 725 Torr = 725/760 atm

Using the ideal gas law: PV = nRT

n = (PV) / (RT) = (0.365 * (725/760)) / (0.08206 * 298)

Now, we have the number of moles of Cl₂(g). Next, we need to find the corresponding moles of MnO₂(s) by using the stoichiometric ratio between MnO₂(s) and Cl₂(g). From the balanced equation, we can see that for every 1 mole of MnO₂(s), 1 mole of Cl₂(g) is produced.

Therefore, the moles of MnO₂(s) = moles of Cl₂(g) = [calculated value above]

Finally, we can convert moles of MnO₂(s) to grams using the molar mass of MnO₂(s). The molar mass of MnO₂(s) is 86.9375 g/mol.

Grams of MnO₂(s) = moles of MnO₂(s) * molar mass of MnO₂(s)