Question

A 0.8690 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 to yield 1.933 g of AgCl. Calculate the percent by mass of each compound in the mixture.

% mass NaCl = ?

% mass KCl = ?

Answer 1

% NaCl = 56.98 %

% KCl = 43.02 %

Explanation

mass AgCl = 1.933 g

moles AgCl = (mass AgCl) / (molar mass AgCl)

moles AgCl = (1.933 g) / (143.32 g/mol)

moles AgCl = 0.01349 mol

Let mass NaCl = x

moles NaCl = (mass NaCl) / (molar mass NaCl)

moles NaCl = (x) / (58.44 g/mol)

moles NaCl = (x / 58.44) mol

moles AgCl produced from NaCl = moles NaCl

moles AgCl produced from NaCl = (x / 58.44) mol

Similarly, moles AgCl produced from KCl = (0.8690 - x) / 74.55 mol

Total moles of AgCl produced = (moles AgCl produced from NaCl) + (moles AgCl produced from KCl)

0.01349 mol = (x / 58.44) mol + (0.8690 - x) / 74.55 mol

Solving for x, x = 0.4952 g

Mass NaCl = x = 0.4952 g

% NaCl = (mass NaCl / mass sample) * 100

% NaCl = (0.4952 g / 0.8690 g) * 100

% NaCl = 56.98 %