Question

A 0.3146-g sample of a mixture of nacl(s) and kbr(s) was dissolved in water. the resulting solution required 55.00 ml of 0.08765 m agno3(aq) to precipitate the cl–(aq) and br–(aq) as agcl(s) and agbr(s). calculate the mass percentage of nacl(s) in the mixture.

Answer 1

The balanced chemical equation representing the reaction of NaCl with AgNO3 is,

$NaCl (aq) + AgNO_{3}(aq) --> AgCl (s) + NaNO_{3}(aq)$

The balanced chemical equation representing the reaction of KBr with AgNO3 is,

$KBr (aq) + AgNO_{3}(aq) --> AgBr(s) + KNO_{3}(aq)$

Moles of $AgNO_{3}$ = $0.08765 \frac{mol}{L} * 55.0 mL * \frac{1 L}{1000 mL} = 0.00482075 mol AgNO_{3}$

0.00482075 mol $AgNO_{3}$ completely removes Chlorides and bromides in the sample.

Mole ratio of $Ag^{+}to Cl^{-} is 1:1$

Mole ratio of $Ag^{+}to Br^{-} is 1:1$

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

$\frac{x}{58.44g/mol} +\frac{y}{119.0g/mol}= 0.00482075 mol$

$\frac{0.3146-y}{58.44g/mol} +\frac{y}{119.0 g/mol} = 0.00482075 mol$

$\frac{119(0.3146-y) + 58.44y}{6954.36}=0.00482075$

$y = 0.0646$

Therefore mass of KBr = 0.0646 g

Mass of NaCl = 0.3146 - 0.0646 g = 0.250 g

Mass % of NaCl in the sample = $\frac{0.250 g NaCl}{0.3146 g sample} * 100 = 79.5 %$