Question

38)What is the experimental yield (in g of precipitate) when 17.8 mL of a 0.6 M solution of sodium chloride is combined with 16.5 mL of a 0.674 M solution of silver nitrate at a 88.5% yield?

Answer 1

Given,

Concentration of sodium chloride solution = 0.6 M

Volume of sodium chloride solution = 17.8 mL x ( 1L /1000 mL) = 0.0178 L

Concentration of silver nitrate solution = 0.674 M

Volume of silver nitrate solution = 16.5 mL x ( 1 L /1000 mL) = 0.0165 L

Calculating the number of moles of NaCl and AgNO₃,

Number of moles of NaCl = 0.6 M x 0.0178 L = 0.01068 mol NaCl

Number of moles of AgNO₃ = 0.674 M x 0.0165 L = 0.01112 mol AgNO₃

Now, the reaction between NaCl and AgNO₃ is,

NaCl(aq) + AgNO₃(aq) AgCl(s) + NaNO₃(aq)

Now, calculating the number of moles of AgCl forming from given moles of NaCl and AgNO₃,

= 0.01068 mol NaCl x ( 1 mol AgCl / 1 mol NaCl)

= 0.01068 mol AgCl

Similarly,

= 0.01112 mol AgNO₃ x ( 1 mol AgCl / 1 mol AgNO₃)

= 0.01112 mol AgCl

The moles of AgCl formed from NaCl are less in quantity thus, NaCl is the limiting reactant.

Thus, using the moles of AgCl calculated from the limiting reactant(NaCl), calculating the theoretical yield of precipitate,

= 0.01068 mol AgCl x (143.32 g / 1 mol)

= 1.53 g of AgCl

Now, we know the formula to calculate percent yield,

Percent yield = [ Experimental yield / Theoretical yield ]x 100

Rearranging the formula,

Experimental yield = [Percent yield /100 ] x Theoretical yield

Substituting the known values,

Experimental yield = [88.5 /100 ] x 1.53 g

Experimental yield = 1.35 g

Thus, the experimental yield of precipitate = 1.4 g [ 2 s.F]