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(Exercise 5.2.14 modified) A common method used to produce bleach (sodium hypochlorite) is by the reaction Cl2 + 2NaOH  NaCl + NaCl + H2O. Chlorine gas is bubbled through an aqueous solution of sodium hydroxide, after which the desired product is separated from the NaCl (by product). A water-NaOH solution that contains 1200 lb of pure NaOH is reacted with 800 lb of gaseous Cl2. The NaOCl formed weighs 600 lb. a. What was the limiting reactant

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Answer:

Chlorine is limiting reactant

Explanation:

Based on the reaction:

Cl₂ + 2NaOH → NaClO + NaCl + H₂O

1 mole of chlorine reacts with 2 moles of NaOH

To find limiting reactant, we need to determine the moles of the reactants:

Moles Cl₂ -Molar mass: 70.9g/mol-:

800lb Cl₂ * (453.6g / 1lb) * (1mol / 70.90g) =

5118 moles Cl₂

Moles NaOH -Molar mass: 40g/mol-:

1200lb NaOH * (453.6g / 1lb) * (1mol / 40g) =

13608 moles NaOH

For a complete reaction of 13608 moles of NaOH you need:

13608 moles NaOH * (1mol Cl₂ / 2 moles NaOH) = 6804 moles of Cl₂

As the solution contains just 5118 moles of chlorine,

Chlorine is limiting reactant

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