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)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

How much MnO2(s)
should be added to excess HCl(aq)
to obtain 105 mL Cl2(g)
at 25 °C and 765 Torr
?

Answer 1

The first step in solving this problem is to use the ideal gas law to calculate the number of moles of Cl2(g) produced:

PV = nRT

where P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the gas constant (0.082 L atm/mol K), and T is the temperature in Kelvin.

Converting the given values to the appropriate units, we get:

P = 765 Torr / 760 Torr/atm = 1.01 atm
V = 105 mL / 1000 mL/L = 0.105 L
T = 25°C + 273.15 = 298.15 K

Substituting these values into the ideal gas law, we get:

n = PV/RT = (1.01 atm)(0.105 L) / (0.082 L atm/mol K)(298.15 K) = 0.00414 mol Cl2(g)

According to the balanced chemical equation, 1 mole of MnO2 reacts with 4 moles of HCl to produce 1 mole of Cl2(g). Therefore, we need 0.00414/4 = 0.00104 mol of MnO2 to produce this amount of Cl2(g).

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

0.00104 mol MnO2 x 86.94 g/mol MnO2 = 0.0907 g MnO2

Therefore, we need approximately 0.0907 g of MnO2 to produce 105 mL of Cl2(g) at 25°C and 765 Torr when reacted with excess HCl(aq).