Answer to Question #179113 in Chemistry for Alex

Question

A 0.8770g sample containing NaCl and KCl was treated with AgNO_3. The AgCl formed had a mass of 1.913g. Calculate the %Na and %K in the sample

Answer 1

Let’s take m(NaCl) = x and m(KCl) = y

The resulting chemical reactions:

x m1

NaCl+ AgNO₃= AgCl+ NaNO₃

58.44g 143.32g

m1 = 143.32 x / 58.44 = 2.45x(g)

y m2

KCl+ AgNO₃= AgCl+ KNO₃

74.55g 143.32g

m2 = 143.32y / 74.55 = 1.92y(g)

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%