Question

a 0.8870 g sample containing only NaCl and KCL was treated with AgNO3. the AgCl formed had a mass of 1.913 g. Calculate the %Na and %K in the sample.

Answer 1

a 0.8870 g sample containing only NaCl and KCL was treated with AgNO3. the AgCl formed had a mass of 1.913 g. Calculate the %Na and %K in the sample.

Solution:

Let's take m(NaCl) = x and m(KCl) = y

The resulting chemical reactions:

x m1

NaCl + AgNO3 = AgCl + NaNO3 m1 = 143.32x / 58.44 = 2.45x(g)

58.44g 143.32g

y m2

KCl + AgNO3 = AgCl + KNO3 m2 = 143.32y / 74.55 = 1.92y(g)

74.55g 143.32g

m1 + m2 = 1.913 g (according to the condition of the problem)

2.45x + 1.92y = 1.913

m(NaCl) + m(KCl) = 0.8870 g (according to the condition of the problem)

x + y = 0.8870

We get the system of equations:

2.45x + 1.92y = 1.913

x + y = 0.8870

Let's solve it:

2.45(0.8870 - y) + 1.92y = 1.913

2.173 - 0.53y = 1.913

y = 0.49

x = 0.887 - 0.49 = 0.397

Let's calculate the %Na and %K in the sample:

%Na = (0.397 / 0.8870) * 100% = 44.76%

%K = (0.49 / 0.8870) * 100% = 55.24%

Answer: %Na = 44.76%; %K = 55.24%

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