Question

A 0.6101 g mixture of KCN (MW = 65.116 g/mol) and NaCN (MW = 49.005 g/mol) was dissolved in water. AgNO3 was added to the solution, precipitating all the CN– in solution as AgCN (MW = 133.886 g/mol). The dried precipitate weighed 1.392 g. Calculate the weight percent of KCN and NaCN in the original sample.

Answer 1

Let Mass of KCN = y

Mass of NaCN = 0.6101 - y

Moles of KCN = mass/molecular weight

= y / 65.116

Moles of NaCN = mass/molecular weight

= ( 0.6101 - y ) / 49.005

Now, KCN -----> K+ + CN-

NaCN -----> Na+ + CN-

Total Moles of CN- = (y / 65.116) + ( 0.6101 - y / 49.005)

moles of AgCN = mass/molecular weight

= 1.392 / 133.886

moles of AgCN = 0.0103969

the reaction is Ag+ + CN- ---> AgCN

Moles of CN- = (y / 65.116) + ( 0.6101 - y / 49.005)

0.0103969 = 0.015357 y + 0.0124497 - 0.020406 y

?0.0020528 = ?0.005049 y

y = 0.40657 g = Mass of KCN

Mass of NaCN = 0.6101 - 0.40657 = 0.20352 g

Wt% of KCN = 0.40657 x 100 / 0.6101 = 66.64%

Wt% of NaCN = 0.20352 x 100/0.6101 = 33.36 %