Determine the quantity (mg/L) of sodium hypochlorite (NaOCl) that would be required to oxidize 0.5 mg/L?

Question

Determine the quantity (mg/L) of sodium hypochlorite (NaOCl) that would be required to oxidize 0.5 mg/L of ferrous ion (Fe2+) according to the following reaction:

2 Fe2+ + NaOCl + 2 H+ 2 Fe3+ + Na+ + H2O + Cl-

Answer 1

I believe that your equation should be:
2 Fe(2+) + NaOCl + 2 H(+) ----> 2 Fe(3+) + Na(+) + H2O + Cl(-)
where the charges are in the parentheses.
To solve this problem, we need:
The atomic mass of Fe: 55.84
The molecular mass of NaOCl: 22.98 + 15.99 + 35.45 = 74.42
Let the quantity of NaOCl needed be x (mg/L).
Since 74.42 g of NaOCl can oxidize 2*55.84 g of Fe(2+), we have:
x/ 74.42 = 0.5/ 111.68
x = 74.42*0.5/ 111.68 = 0.33 (mg/L)
Thus, 0.33 mg/L of sodium hypochlorite (NaOCl) is required to oxidize 0.5 mg/L of ferrous ion (Fe2+)