Reaction 2 occurs when an excess of 6 M HCl (aq)solution is added to 100 mL of NaOCl (aq) of unknown concentration.

If the reaction goes to completion and 0.010 mol of Cl2(g)is produced, then what was the molarity of the NaOCl solution?

(A)0.0010 M

(B)0.010 M

(C)0.10 M

(D)1.0 M

[NaOCl] = 0.1M   (option c.)

Explanation:

We determine the reaction:

NaOCl(aq) + HCl (aq) →NaOH (aq) +  Cl₂(g)

Ratio is 1:1, so if I produce 0.01 moles of chlorine I used 0.01 mol  of hypochlorite.

Our volume of solution by water (solvent) and NaOCl (solute) was 100 mL, so in order to determine molarity we divide moles by volume (L)

100 mL . 1L/1000mL = 0.1L

Molarity → 0.01 mol /0.1L = 0.1M