To obtain 345 kCl at 25°C and 725 torr, approximately 990 grams of MnO2 should be added to excess HCl. This calculation is based on stoichiometry and the ideal gas law, which allow us to determine the number of moles of Cl2 produced and then convert it to the required mass of MnO2.
To determine how much MnO2 should be added to excess HCl to obtain 345 kCl at 25°C and 725 torr, we need to use stoichiometry and the ideal gas law.
1. Write the balanced chemical equation for the reaction between MnO2 and HCl:
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
2. Use the ideal gas law equation to calculate the number of moles of Cl2 produced:
PV = nRT
n = PV / RT
n = (725 torr) * (345 kCl) / [(0.0821 L·atm/mol·K) * (298 K)]
n ≈ 11.4 moles of Cl2
3. From the balanced chemical equation, we can see that 1 mole of MnO2 produces 1 mole of Cl2. Therefore, the number of moles of MnO2 needed is also approximately 11.4 moles.
4. To calculate the mass of MnO2 needed, we need to know its molar mass, which is 86.94 g/mol.
Mass of MnO2 = moles of MnO2 * molar mass of MnO2
Mass of MnO2 = 11.4 moles * 86.94 g/mol
Mass of MnO2 ≈ 990 grams
Therefore, approximately 990 grams of MnO2 should be added to excess HCl to obtain 345 kCl at 25°C and 725 torr.