(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

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,

wher is the rest of the question