Question

Bleach (sodium hypochlorite, NaOCl) is produced by the reaction of a sodium hydroxide solution with chlorine gas. Water and NaCl are also produced in the reaction.

1.75 liters of a 45% NaOH solution (S.G = 1.478) is reacted with 275 liters of gaseous Cl2 (at STP). The product is analyzed and it is determined that 618 g of NaOCl was formed. Determine:

The limiting reactant.

The percent excess of the excess reactant, if any.

The conversion.

The yield of NaOCl.

Answer 1

Ans. # Step 1: Calculate moles of Cl₂:

Molar volume at STP = 22.4 L/ mol

So,

            Moles of Cl₂ = Volume of Cl₂ / Molar volume at STP

                                    = 275 L / (22.4 L / mol)

                                    = 12.28 mol

# Step 2: Calculate moles of NaOH:

Given, [NaOH] = 45% (w/w)           [Note: since SG of solution is given, [NaOH] shall be in terms of % w/w]

            Volume of solution = 1.75 L = 1750.0 mL

            Specific gravity of solution = 1.478

            Or, density of solution = 1.478 g/ mL = 1.478 kg/ L

Now,

            Mass of solution = Volume of solution x Density

                                                = 1.75 L x (1.478 kg/ L)

                                                = 2.5865 kg

# Mass of NaOH = 45 % of mass of solution

                                    = 45% of 2.5865 kg

                                    = 1.163925 kg

                                    = 1163.925 g

# Moles of NaOH = Mass/ Molar mass

                                    = 1163.925 g / (40.0 g/ mol)

                                    = 29.09 mol

# Step 3: Determine limiting reactant:

Balanced Reaction: 2NaOH + Cl₂ → NaCl + NaOCl + H₂O

Theoretical molar ratios of reactants = NaOH : Cl₂ = 2 : 1

Experimental molar ratios of reactants = NaOH : Cl₂ = 29.09 mol : 12.28 mol = 2.37 : 1

Comparing the theoretical and experimental molar ratios, the moles of NaOH is greater than theoretical value of 2.0 moles while that of Cl₂ is kept constant at 1.0 mol.

So,

            NaOH is the reagent in excess.

            Cl₂ is the limiting reactant.

# Step 4: Determine yield:

According to the stoichiometry of balanced reaction, 2 mol NaOH forms 1 mol NaOCl.

# Formation of product follows the stoichiometry of limiting reactant.

So,

Theoretical moles of NaOCl produced = (1/2) x moles of NaOH

                                                = (1/2) x 29.09 mol

                                                = 14.545 mol

Theoretical mass of NaOCl produced = Theoretical moles x Molar mass

                                                = 14.545 mol x (74.441868 g/mol)

                                                = 1082.76 g

Now,

            % Yield of NaOCl = (Actual yield / Theoretical yield) x 100

                                                = (618 g/ 1082.76 g) x 100

                                                = 57.08 %