Question

a 1.0g sample of a mixture which contains only NaCl and KCl have a precipitate of AgCl which weighed 2.0 g. what are the percentage of Na and K in the mixture?

Answer 1

Solution.

$n(AgCl) = \frac{m}{M} = \frac{2.0}{143.32} = 0.014 \ mol$

$Cl^- + Ag^+ = AgCl$

x = n(NaCl)

y = n(KCl)

$1) x + y = 0.014$

$2) 58.44 \times x + 74.55 \times y = 1.0$

x = n(NaCl) = 0.003 mol

y = n(KCl) = 0.011 mol

$m(NaCl) = n(NaCl) \times M(NaCl) = 0.180 \ g$

$m(KCl) = n(KCl) \times M(KCl) = 0.820 \ g$

$w(NaCl) = 18 \%$

$w(KCl) = 82 \%$

$m(Na) = \frac{m(NaCl) \times w(Na \ in \ NaCl)}{100 \%}$

$m(K) = \frac{m(KCl) \times w(K \ in \ KCl)}{100 \%}$

m(Na) = 0.071 g

m(K) = 0.430 g

$w(Na \ in \ mix) = \frac{m(Na)}{m(mix.)} = 7.10 \%$

$w(K \ in \ mix) = \frac{m(K)}{m(mix.)} = 43.00 \%$

Answer:

w(Na in mix) = 7.10 %

w(K in mix) = 43.00 %