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 (mol)/(L) * 55.0 mL * (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

`(x)/(58.44g/mol) +(y)/(119.0g/mol)= 0.00482075 mol`

`(0.3146-y)/(58.44g/mol) +(y)/(119.0 g/mol) = 0.00482075 mol`

`(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 =
`(0.250 g NaCl)/(0.3146 g sample) * 100 = 79.5 %`