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 185 mL Cl2(g) at 25°C and 785 Torr?
MnO2(s) + 4HCl(aq)⟶MnCl2(aq) + 2H2O(l) + Cl2(g)
You want 185 mL Cl2 @ the conditions listed. Convert to moles.
Use PV = nRT and solve for n. You have P which is 785 torr/760 = ? atm.
R = 0.08205; T = 25 C + 273 = 298 K and V = 0.185 L. From the equation you see for every 1 mol Cl2 obtained you must start with 1 mole MnO2; therefore, mols Cl2 from the first calculation = mols MnO2 needed initially.
Then convert to grams. grams MnO2 = mols MnO2 x molar mass MnO2 = ?
Post your work if you get stuck.