Question

chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid HCL as described by the chemical equation MnO2(S)+4HCl(aq)-->MnCl2 +2H2)+Cl2 how much MnO2 should be added to excess HCl to obtain 345 kCl at 25 C and 725 torr

Answer 1

To obtain 345 kCl of Cl2 gas at 25 C and 725 torr, 2.77 g of MnO2 should be added to excess HCl.

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 the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for n, which is the number of moles of Cl2 gas produced in the reaction. Then, we can use stoichiometry to determine the number of moles of MnO2 needed to produce that amount of Cl2 gas.

First, we need to convert the given pressure and temperature to the appropriate units. 725 torr is equivalent to 0.960 atm, and 25 C is equivalent to 298 K.

Using the ideal gas law equation, we can solve for n:

n = PV/RT = (0.960 atm)(0.345 L)/(0.0821 L·atm/mol·K)(298 K) = 0.0159 mol Cl2

According to the balanced chemical equation, 1 mole of MnO2 produces 1/2 mole of Cl2. Therefore, we need twice as many moles of MnO2 as we do Cl2:

n(MnO2) = 2 × n(Cl2) = 2 × 0.0159 mol = 0.0318 mol

Finally, we can use the molar mass of MnO2 to convert moles to grams:

m(MnO2) = n(MnO2) × M(MnO2) = 0.0318 mol × 86.94 g/mol = 2.77 g

Therefore, we need 2.77 g of MnO2 to react with excess HCl to produce 345 kCl of Cl2 gas at 25 C and 725 torr.