If 30.0 mL of 0.150 M CaCl2 is added to 35.0 mL of 0.100 M AgNO3,...

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?

Expert Solution

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

queen_honey_blossom answered 2 years ago

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