Question

When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation

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

Part A

What mass of silver chloride can be produced from 1.28 L of a 0.267 M solution of silver nitrate?

Express your answer with the appropriate units.

Part B

The reaction described in Part A required 3.60 L of calcium chloride. What is the concentration of this calcium chloride solution?

Express your answer with the appropriate units

Answer 1

Part A

no. of mole = molarity volume of solution in liter

no. of mole of silver nitrate = 0.267 1.28 = 0.34176 mole

According to reaction 2 mole of silver nitrate produce 2 mole of silver chloride then 0.34176 mole of silver nitrate produce 0.34176 mole of silver chloride

molar mass of silver chloride = 143.32 gm / mole then 0.34176 mole of silver chloride = 143.32 0.34176 = 48.981 gm of silver chloride.

48.981 gm silver chloride produced.

Part B

According to reaction 2 mole of silver nitrate react with 1 mole of calcium chloride then to react with 0.34176 mole of silver nitrate required calcium chloride = 0.34176 / 2 = 0.17088 mole of calcium chloride

Molarity = no. of mole/volume of solution in liter

Molarity of calcium cloride = 0.17088 / 3.60 = 0.0475 M

Molarity of calcium cloride = 0.0475 M