Question

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?

Answer 1

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