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)

What mass of silver chloride can be produced from 1.51 L of a 0.201 M solution of silver nitrate?
Express your answer with the appropriate units.

The reaction described in Part A required 3.51 L of calcium chloride. What is the concentration of this calcium chloride solution?

Express your answer with the appropriate units.

Answer 1

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

Number of moles of AgNO₃ is , n = Molarity x volume in L

                                                 = 0.201 M x 1.51 L

                                                 = 0.3035 moles

According to the above balanced equation ,

2 moles of AgNO₃ produces 2 moles of AgCl

0.3035 moles of AgNO₃ produces 0.3035 moles of AgCl

Molar mass of AgCl = At.mass of Ag + At.mass of Cl

                              = 107.9 + 35.5

                              = 143.4 g/mol

So mass of AgCl produced , m = number of moles x molar mass

                                              = 0.3035 mol x 143.4 (g/mol)

                                              = 43.522 g

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According to the above balanced equation ,

2 moles of AgNO₃ reacts with 1 mole of CaCl₂

0.3035 moles of AgNO₃ reacts with 0.3035/2 = 0.152 mole of CaCl₂

So Molarity of CaCl₂ solution , M = number of moles / Volume in L

                                                 = 0.152 mol / 3.51L

                                                 = 0.043 M