Aqueous solutions of sodium hypoch lorite (NaOCI), best known as bleach, are prepared by the reaction of sodium hydroxide with chlorine: 2 NaOH (aq) Cl2 (g) - >NaOCI (aq) + H20 (I) + NaCl (aq) How many grams of NaOH are needed to react with 25.0 g of chlorine?

28.2 g of NaOH

Explanation:

We need to calculate the grams of NaOH needed to react with 25.0 g of Cl₂ in the following reaction:

2 NaOH (aq) + Cl₂ (g) → NaOCI (aq + H₂0 (I) + NaCl (aq)

We are going to solve this by making use of the molar ratio between Cl₂ and NaOH given by the reaction equation where we see that every mol of Cl₂ will react with 2 moles of NaOH.

So first we need to convert the 25.0 g of Cl₂ to moles:

Molar Mass of Cl₂ = 2 x 35.45 = 70.90 g/mol Moles of Cl₂ = 25.0 g / 70.90 g/mol = 0.3526 moles

Then we need to calculate the moles of NaOH needed to react with these moles of Cl₂ knowing that every mol of Cl₂ will react with 2 moles of NaOH:

moles of NaOH = 2 x moles of Cl₂ = 2 x 0.3526 moles = 0.7052 moles

Next we must convert these moles to grams:

Molar Mass of NaOH = 22.990 + 15.999 + 1.008 = 40.00 g/mol Mass of NaOH = 0.7052 moles x 40.00 g/mol = 28.2 g

28.2 g are needed to react with 25.0 g of Cl₂ in the production of NaOCl