Question

A 0.7257 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 1.650 g. Calculate the weight percent of KCN and NaCN in the original sample.

Answer 1

Solution-Let,us assume

Mass of KCN = y

Mass of NaCN = 0.7257 - y

Moles of KCN = y / 65.116 moles

Moles of NaCN = ( 0.7257 - y ) / 49.005

Now,

KCN -----> K+ + CN-

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

Moles of CN- from KCN = Moles of KCN = y / 65.116

Moles of CN- from NaCN = Moles of NaCN = ( 0.7257 - y ) / 49.005

Total moles of CN- = y / 65.116 + ( 0.7257 - y ) / 49.005

AgNO3 + CN- -------> AgCN + NO3-

Moles of AgCN = Moles of CN-

Mass of AgCN = Moles of AgCN * Molar mass

1.650 = [ y / 65.116 + ( 0.7257 - y ) / 49.005 ] * 133.886

0.0123 = [ y / 65.116 + ( 0.7257 - y ) / 49.005 ]

0.0123= 0.0154 * y+0.0148-0.0204*y

0.0123 = 0.0154 * y + 0.0146 - 0.0204 * y

0.0204y - 0.0154y = 0.0146 - 0.0123

0.005 y = 0.0023

y = 0.46 g

Mass of KCN = 0.46 g and Mass of NaCN = 0.7257 - 0.46 = 0.2657 g

Weight % of KCN = ( Mass of KCN / Mass of mixture ) * 100

Weight % of KCN = 0.46 / 0.7257 * 100 = 63.39%

Weight % of NaCN = 0.2657 / 0.7257 * 100 = 36.61%