Question

1. Household bleach (NaOCl) is really a solution of sodium hypochlorite in water. It is made by reacting chlorine gas and sodium hydroxide, as shown in the unbalanced chemical equation below: NaOH (aq) + Cl2 (g) → NaOCl (aq) + NaCl (aq) + H2 O (l) A student bubbles 83.0g of chlorine gas into an excess of aqueous sodium hydroxide. After analyzing his products, he determines that 22.0 grams of bleach are actually made. (a) What is the oxidation number of chlorine in bleach (NaOCl)? (b) Is this reaction above redox? Explain. (c) What is the percent yield of this reaction?

Answer 1

This is a three-part question

Answers:

• a) The oxidation number of chlorine in bleach (NaOCl) is +1.

• b) Yes, this is a redox reaction

• c) The percen yield is 56.8%

Explanation:

1) Part (a) What is the oxidation number of chlorine in bleach (NaOCl)?

a) Rule one: In a neutral compound the sum of the oxidation states is zero.

Since NaOCl is a neutral compound the sum of the oxidation states of Na, O and Cl is 0.

b) Rule two: since Na is an alkaline metal, its oxidation state is +1

c) Rule three: the most common state of oxygen, except in peroxides, is -2.

Then,

• Na: +1
• O:   -2
• Cl:    x

      Sum = +1 - 2 + x = 0 ⇒ x = 2 - 1 = 1

Conclusion: the oxidation state of NaOCl is +1.

2) Part (b) Is this reaction above redox?  

In a redox reaction the oxidation states of some substances increase (get oxidized) and the oxidation states of some substances decrease (get reduced).

The reaction is represented by the chemical equation given:

• NaOH (aq) + Cl₂ (g) → NaOCl (aq) + NaCl (aq) + H₂O (l)

Since the chlorine gas (Cl₂) is a molecule of only chlorine atoms, its oxidation state is zero, and since chlorine is forming compounds on the right side (NaOCl and NaCl) you can immediately conclude that the oxidation state of chlorine changed, and this is a redox reaction.

In fact:

• Oxidation state of Cl in Cl₂: 0

• Oxidation state of Cl in NaOCl: +1 (previously determined)

• Oxidation state of Cl in NaCl: -1 (becasue Na has oxidation state +1 and so +1 - 1 = 0).

Therefore, chlorine is being oxidized (its oxidation state increases from 0 to +1) and is also being reduced (its oxidation state is reduced from 0 to -1), and this is a redox reaction.

3) Part (c) What is the percent yield of this reaction?

a) Chemical equation (given)

• NaOH (aq) + Cl₂ (g) → NaOCl (aq) + NaCl (aq) + H₂O (l)

b) Theoretical mole ratio:

• 2 mol Cl₂ (g) : 1 mol  NaOCl (aq)

c)  Convert 83.0 g of chlorine gas to moles:

• moles = mass in grams / molar mass

• molar mass of Cl₂(g) = 79.906 g/mol

• moles = 83.0 g / 79.906 g/mol = 1.039 mol Cl₂(g)

d) Determine the theoretical yiedl using proportions:

• x / 1.039 mol Cl₂ (g) = 1 mol NaOCl / 2 mol Cl₂ (g)

• ⇒ x = 0.5195 mol NaOCl

e) Convert 0.5195 mol NaOCl to grams:

• molar mass NaOCl = 22.99 g/mol + 16.00 g/mol + 35.453 g/mol = 74.443 g/mol

• mass = number of mol × molar mass = 0.5195 mol × 74.443 g/mol = 38.67 g

         That must be rounded to 3 significant figures (such as the mass of belach is given: 22.0 g)

• Theoretical yiedl of NaOCl = 38.7 g

f) Calculate the percent yiled:

• Percent yield = (actual yield / theoretical yield) × 100

• Percent yield = (22.0 g / 38.7 g) × 100 = 56.8%