How many grams of silver chloride can be prepared by the reaction of 105.0 mL of

Question

How many grams of silver chloride can be prepared by the reaction of 105.0 mL of 0.21 Msilver nitrate with 105.0 mL of 0.14 M calcium chloride?
g

Calculate the concentrations of each ion remaining in solution after precipitation is complete.

Ag+ M NO3? M Ca2+ M Cl? M

How many grams of silver chloride can be prepared by the reaction of 105.0 mL of 0.21 Msilver nitrate with 105.0 mL of 0.14 M calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.

Explanation / Answer

Approximate MW's
Ag = 107.9
NO3 = 14 + 48 = 62; AgNO3 = 16.9
Ca = 40
Cl = 35.5; CaCl2 = 111

The reaction is:

2AgNO3(aq) + CaCl2(aq) => Ca(NO3)2(aq) + 2AgCl(s)?
2 moles of AgNO3 react with 1 mole of CaCl2

105 ml of 0.21M AgNO3 contains 0.21 x105/1000 = 0.02205 moles AgNO3

105 ml of 0.14M CaCl2 contains 0.14 x105/1000 = 0.0147 moles

The CaCl2 is in excess, so all the Ag is in the precipitate

The mass of Ag originally present in 0.02205 moles AgNO3 =
0.02205 x 107.9 = 2.379 g

Mass of AgCl precipitate = 0.02205 x 143.4 =3.162 g

The solubility of AgCl2 in H2O is very small (52

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