Question

1. Calculate the theoretical yield (in grams) of the precipitate that should form in Test Tubes B and C. (assume, 20 drops = 1. 0mL)

2. 	B	C
   Mass of empty test tube (g)	12.358g	7.675g
   Drops of NiCl2	30	24
   Drops of AgNO3	20	32
   Conductivity of solution (μS/cm)	21,521 (μS/cm)	21,005 (μS/cm)
   Mass of precipitate and test tube (g)	12.415 g	7.805 g

3. Tube C:

4. Concentrations .40 M Nicl2

   .60 M Ag NO3

Answer 1

Reaction : NiCl2 (aq) + 2 AgNO3 (aq) ====> 2 AgCl (s) + Ni(NO3)2 (aq)

theoretical yield (in grams) of the precipitate

Tube B :

20 drops = 1. 0mL

moles of AgNO3 = 1.0 mL* 1 L/1000 mL* 0.60 mol/L = 0.0006 moles

moles of NiCl2 =   1.5 mL* 1 L/1000 mL* 0.40 mol/L = 0.0006 moles

this is 2 : 1 reaction, thus limiting reagent = AgCl

moles of AgCl = 0.0006 mol

mass of ppt (AgCl) = 143.32 g/mol* 0.0006 moles = 0.0860 g

Tube C :

20 drops = 1. 0mL

moles of AgNO3 = 1.6 mL* 1 L/1000 mL* 0.60 mol/L = 0.00096 moles

moles of NiCl2 =   1.2 mL* 1 L/1000 mL* 0.40 mol/L = 0.00048 moles

this is 2 : 1 reaction, thus both reactant are in required equivalent ratio.

moles of AgCl = 0.00096 mol

mass of ppt (AgCl) = 143.32 g/mol* 0.00096 moles = 0.1376 g