Answer to Question #309559 in Chemistry for Zel

Question

Treatment of a 0.2500g sample of impure potassium chloride with an excess of AgNO3 resulted in the formation of 0.2912 g of AgCl (74.55 g/mol). Calculate the percentage of KCl (143.32 g/mol) in the sample

Answer 1

Solution:

Calculate the moles of AgCl:

The molar mass of AgCl is 143.32 g/mol

Therefore,

Moles of AgCl = (0.2912 g AgCl) × (1 mol AgCl / 143.32 g AgCl) = 0.002032 mol AgCl

Balanced chemical equation:

KCl(aq) + AgNO₃(aq) → AgCl(s) + KNO₃(aq)

According to stoichiometry:

1 mol of KCl produces 1 mol of AgCl

X mol of KCl produces 0.002032 mol of AgCl

Thus,

Moles of KCl = X = (0.002032 mol AgCl) × (1 mol KCl / 1 mol AgCl) = 0.002032 mol KCl

Calculate the mass of KCl:

The molar mass of KCl is 74.55 g/mol

Therefore,

Mass of KCl = ( 0.002032 mol KCl) × (74.55 g KCl / 1 mol KCl) = 0.151485 g KCl = 0.1515 g KCl

Calculate the percentage of KCl in the sample:

%KCl = (Mass of KCl / Mass of sample) × 100%

%KCl = (0.1515 g / 0.2500 g) × 100% = 60.6%

%KCl = 60.6%

Answer: The percentage of KCl in the sample is 60.6%