Question

Part A What mass of silver chloride can be produced from 1.46 L of a 0.282 M solution of silver nitrate? Express your answer with the appropriate units. Part B The reaction described in Part A required 3.01 L of magnesium chloride. What is the concentration of this magnesium chloride solution? Express your answer with the appropriate units.

Answer 1

Sol.

Reaction :

MgCl2 + 2AgNO3 ----> Mg(NO3)2 + 2AgCl

Part A :

moles of silver nitrate , AgNO3

= conc. of AgNO3 × Volume of AgNO3  

= 0.282 × 1.46

= 0.41172 mol  

From reaction ,

moles of silver chloride , AgCl = moles of AgNO3 = 0.41172 mol

As molar mass of AgCl = 143.32 g / mol

So , mass of AgCl produced

= 0.41172 × 143.32  

=   59.008 g

Part B :

From reaction , moles of magnesium chloride , MgCl2

= moles of AgNO3 / 2

= 0.41172 / 2   

= 0.20586 mol

Also , Volume of MgCl2 = 3.01 L

So , Conc. of MgCl2  

= 0.20586 / 3.01

=   0.068 M