The balanced chemical equation representing the reaction of NaCl with AgNO3 is,
[tex]NaCl (aq) + AgNO_{3}(aq) --> AgCl (s) + NaNO_{3}(aq)[/tex]
The balanced chemical equation representing the reaction of KBr with AgNO3 is,
[tex]KBr (aq) + AgNO_{3}(aq) --> AgBr(s) + KNO_{3}(aq)[/tex]
Moles of [tex]AgNO_{3}[/tex] = [tex]0.08765 \frac{mol}{L} * 55.0 mL * \frac{1 L}{1000 mL} = 0.00482075 mol AgNO_{3}[/tex]
0.00482075 mol [tex]AgNO_{3}[/tex] completely removes Chlorides and bromides in the sample.
Mole ratio of [tex]Ag^{+}to Cl^{-} is 1:1[/tex]
Mole ratio of [tex]Ag^{+}to Br^{-} is 1:1[/tex]
So, the total moles of chloride and bromide in the sample = 0.00482075 mol
Let mass of NaCl be x g
Mass of KBr be y g
Total mass of sample = 0.3146 g
=> x + y = 0.3146 g
x = 0.3146 -y
Total number of moles of NaCl + KBr = 0.00482075 mol
[tex]\frac{x}{58.44g/mol} +\frac{y}{119.0g/mol}= 0.00482075 mol[/tex]
[tex]\frac{0.3146-y}{58.44g/mol} +\frac{y}{119.0 g/mol} = 0.00482075 mol[/tex]
[tex]\frac{119(0.3146-y) + 58.44y}{6954.36}=0.00482075[/tex]
[tex]y = 0.0646[/tex]
Therefore mass of KBr = 0.0646 g
Mass of NaCl = 0.3146 - 0.0646 g = 0.250 g
Mass % of NaCl in the sample = [tex]\frac{0.250 g NaCl}{0.3146 g sample} * 100 = 79.5 %[/tex]
