Answer to Question #99123 in General Chemistry for m

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) + 4 HCl(aq)= MnCl2(aq) + 2 H2O(l)+Cl2(g)

How much MnO2 should be added to excess HCl(aq) to obtain 245mL Cl2(g) at 25 °C and 705 Torr ?

Answer 1

From the combined gas law:

$$V_2(Cl_2) = {T_2p_1V_1(Cl_2) \over T_1p_2}$$

$$V_2(Cl_2) = {273.15*705*245 \over 298.15*760} = 208 mL$$

$$n(Cl_2) = {0.208 \over 22.4} = 0.009 mol$$

From the stoichiometric reaction relationships:

$$n(MnO_2) = n(Cl_2) = 0.009mol$$

$$m(MnO_2) = n*M = 0.009*86.937 = 0.782g$$

Answer: 0.782 g