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.
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?
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1 answer