Question

if 30.0 ml of 0.150M CaCL2 is added to 15 ml of 0.100M AgNO3. what is the mass in grams of AgCL precipitate?

Answer 1

CaCl2(aq) + 2AgNO3(aq) ⎯⎯→ Ca(NO3)2(aq) + 2AgCl(s)

30.0 mL of 0.150 M 15.0 mL of 0.100 M ? g

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Ag+ and Cl– combine in a 1:1 mole ratio to produce AgCl.

Let’s calculate the amount of Ag+ and Cl– in solution.

mol Ag+ = 15.0 mL soln * ( 0.100 mol Ag+/ 1000 mL soln ) = 1.50 × 10^−3 mol Ag+

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mol Cl− = 30.0 mL soln * ( 0.150 mol CaCl2 /1000 mL soln )* (2 mol Cl^− 1 / mol CaCl2 )

= 9.00 × 10^−3 mol Cl^−

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Since Ag+ and Cl– combine in a 1:1 mole ratio,

AgNO3 is the limiting reagent.

Only 1.50 × 10^−3 mole of AgCl can form.

Converting to grams of AgCl:

1.50 × 10^−3 mol AgCl * ( 143.4 g AgCl /1 mol AgCl )

=0.215 g AgCl-------------------answer