Question

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide: 4HCL(aq) +MnO2(s) ----> MnCl2(aq) + H2O(l) + Cl2(g) You add 34.9 g of MnO2 to a solution containing 46.3 g of HCl.

Answer 1

4HCl(aq) +MnO2(s) ----> MnCl2(aq) + H2O(l) + Cl2(g)

We need to know the molar mass of all the reactants and products to calculate the chlorine gas produced.

Reaction takes place in stoichiometric ratio based on the balanced reaction.

Molar Mass of HCl = 36.46 g/mol

Molar Mass of MnO2 = 86.93 g/mol

Molar mass of Cl2 = 70.90 g/mol

Therefore, for 34.9 g of MnO2, HCl required = 4*36.46*34.9/86.93 = 58.55 g but HCl used was 46.3 g. So, HCl is the limiting reagent.

Now 4*36.46 g of HCl produces 70.90 g of Cl2 gas.

46.30 g of HCl will produce 70.90*46.30/(4*36.46) g = 22.50 g of Cl2 gas will be produced.