Question

Calculate how many grams of sodium hypochlorite can be theoretically formed when 1.23 mol of sodium hydroxide is combined with 1.36 mol of chlorine.

Answer 1

45.78 g NaClO

Step-by-step explanation:

The reaction that takes place is:

• 2NaOH + Cl₂ → NaCl + NaClO + H₂O

In order to react completely, 1.36 moles of chlorine would require (2*1.36) 2.72 moles of NaOH. There are more moles than that, so NaOH is the limiting reactant.

We calculate the moles of NaClO formed, from the limiting reactant:

• 1.23 mol NaOH *
  `(1molNaClO)/(2molNaOH)` = 0.615 mol NaClO

Finally we convert NaClO moles to grams, using its molecular weight:

• 0.615 mol NaClO * 74.44 g/mol = 45.78 g NaClO