Question #114575

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?

Expert's answer

**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 %

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