Question

If 30.0 mL of 0.150 M CaCl2 is added to 35.0 mL of 0.100 M AgNO3, what is the mass of the AgCl precipitate?

Answer 1

moles of CaCl2 = 30 x 0.150 / 1000 = 4.5 x 10^-3

moles of AgNO3 = 35 x 0.1 / 1000 = 3.5 x 10^-3

CaCl2 +      2 AgNO3   ------------------> 2 AgCl +   Ca(NO3)2

    1             2                                         2

4.5 x 10^-3       3.5 x 10^-3                       ?

here limiting reagent AgNO3 . so AgCl formed based on that.

2 mol of AgNO3 ---------------   2 mol of AgCl

3.5 x 10^-3 AgNO3 -------------> ??

moles of AgCl = 3.5 x 10^-3 x 2 / 2 = 3.5 x 10^-3

mass of AgCl = moles x molar mass

                     = 3.5 x 10^-3 x 143.3

                    = 0.5 g

mass of AgCl = 0.5 g