Question

CaCl2 reacts with AgNO3 in aqueous solutions to produce a precipitate of AgCl that can be filtered and weighed. The balanced equation for the reaction is:

CaCl2 (aq) + 2 AgNO3 (aq) → 2 AgCl (s) + Ca(NO3)2 (aq)
Suppose you have a mixture that contains CaCl2, plus other compounds that do not react with AgNO3. If 0.2918

g of the mixture yields 0.4462 g of AgCl, what is the percentage of CaCl2 in the mixture? Molar masses: CaCl2 110.98 g/mol AgCl 143.32 g/mol

Answer 1

CaCl2 (aq) + 2 AgNO3 (aq) → 2 AgCl (s) + Ca(NO3)2 (aq) Here 1 mole Cacl2 = 110.98 g

1 mole 2 mole 1 mole AgCl = 143.32 g

1 mole CaCl2 will give 2 mole of AgCl

110.98 g of CaCl2 will give (2 x 143.32 ) g of AgCl

? g of CaCl2 will give 0.4462 g of AgCl

  

= 110.98 g x 0.4462 g / (2 x 143.32) g

= 0.173 g of CaCl2

Therefore percentage of CaCl2 = ( weight of CaCl2 / weight of the mixture ) x 100

= ( 0.173 g / 0.4462 g ) x 100

= 38.77 %