The solution would be like this for this specific problem:
4 NaOCl + S2O3{2-} + 2
OH{-} → 2 SO4{2-} + H2O + 4 NaCl
<span>(0.00456 L) x (0.100 mol/L
S2O3{2-}) x (4 mol NaOCl / 1 mol S2O3{2-}) x (100.0 mL / 25 mL) x </span><span>
<span>(74.4422 g NaClO/mol) = 0.54313 g </span></span>
<span>(5.00 mL) x (1.08 g/mL) =
5.40 g solution </span>
(0.54313 g) / (5.40 g) =
0.101 = 10.1%
So, the average percent by mass of NaClO in the
commercial bleach is 10.1%.
Answer:
The first one
Explanation:
I can't read them well: it should be 
** this is the expanded version
please have look at Periodic table , you will solve it yourself !
