Question

# Silver nitrate reacts with magnesium chloride to form solid silver chloride and aqueous magnesium nitrate. Determine...

Silver nitrate reacts with magnesium chloride to form solid silver chloride and aqueous magnesium nitrate.

Determine the percent yield if a student isolates 58.9 mg of solid silver chloride from the mixture of 0.168 g of silver nitrate and 0.543 g of magnesium chloride.

What mass, in grams, of excess reactant remains?

The reaction is given as :

2AgNO3 + MgCl2 = 2AgCl + Mg(NO3)2

0.168 g of silver nitrate = 0.168 / 169.87 = 9.89 x 10-4 moles

0.543 g of magnesium chloride = 0.543/ 95.211 = 0.0057 moles or 5.7 x 10-3 moles

1 mole of MgCl2 requires 2 moles of AgNO3 , so 5.7 x 10-3 moles will require 2 x 5.7 x 10-3 = 1.14 x 10-2 moles

So clearly AgNO3 is the limiting reagent.

9.89 x 10-4 moles will make 9.89 x 10-4 moles of AgCl

So the theoretical yield of AgCl = 9.89 x 10-4 x 143.32 = 0.142 g

Percent yield = (0.0589 / 0.142) x 100

= 41.48 %

Number of moles of MgCl2 required = (9.89 x 10-4 ) / 2 = 4.945 x 10-4 moles.

Excess MgCl2 = (5.7 x 10-3) - (4.945 x 10-4)

= 5.2 x 10-3 g