help with chem problem

Question

A 0.52970.5297 g mixture of KCNKCN (MW=65.116 g/molMW=65.116 g/mol) and NaCNNaCN (MW=49.005 g/molMW=49.005 g/mol) was dissolved in water. AgNO3AgNO3 was added to the solution, precipitating all of the CN−CN− in solution as AgCNAgCN (MW=133.886 g/molMW=133.886 g/mol). The dried precipitate weighed 1.2061.206 g. Calculate the weight percent of KCNKCN and NaCNNaCN in the original sample.

KCN wt%=

NaCN wt%=

Answer 1

KCN(aq) + AgNO₃(aq) ==> AgCN(s) + KNO₃(aq)

NaCN(aq) + AgNO₃(aq) ==> AgCN(s) + NaNO₃(aq)

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KCN(aq) + NaCN(aq) + 2AgNO₃(aq) ==> 2AgCN(s) + KNO₃(aq) + NaNO₃(aq) .. balanced equation

moles AgCN formed = 1.206 g x 1 mol / 133.886 g = 0.009008 moles AgCN

moles KCN + moles NaCN = 0.009008 moles

let x = g KCN

0.5297 - x = g NaCN

(x / 65.116) + (0.5297 - x / 49.005) = 0.009008 moles

0.01536x + 0.0108 - 0.0204x = 0.009008

-5.046x10^-3x = -1.72x10^-3 (note: see comments and edited version)

x = 0.341 g KCN

0.5297 - 0.341 = 0.189 g NaCN

(be sure to check the math)

KCN wt% = 0.341 g / 0.5297 g (x100%) = 64.4% (note: see comments and edited version)

NaCN wt% = 0.189 g / 0.5297 g (x100%) = 35.7% (note: see comments and edited version)

EDITED:

-5.046x10^-3x = -1.792x10^-3

x = 0.3551 g KCN

0.5297 g - 0.3551 g = 0.1746 g NaCN

KCN wt% = 0.3551 g / 0.5297 g (x100%) = 67.04%

NaCN wt% = 0.1746 g / 0.5297 g (x100%) = 32.96%