Question

A chemist dissolves 0.401 g of silver nitrate in a beaker of water and 0.253 g of magnesium chloride in a second beaker of water. She then mixes the two solutions and isolates a white precipitate. She finds its mass to be 0.292 g. What is her percent yield for the reaction?

Answer 1

Answer on Question 76561 in General Chemistry

.m(AgNO₃)=0.401 g

.m(MgCl₂)=0.253 g

.m(AgCl)=0.292 g

percent yield for the reaction=?

Solution:

Write the equation of the reaction

$$2 \mathrm{AgNO_3} + \mathrm{MgCl_2} = 2 \mathrm{AgCl} + \mathrm{Mg(NO_3)_2}$$

Find the amount of substance

.n(AgNO₃) = $\frac{m(AgNO_3)}{Mr}$ = $\frac{0.401}{170}$ = 0.002 mol

Mr(AgNO₃)= Ar(Ag)+Ar(N)+3 Ar(O)=108+14+3×16=170

.n(MgCl₂) = $\frac{m(MgCl_2)}{Mr}$ = $\frac{0.253}{95}$ = 0.003

Mr(MgCl₂)= Ar(Mg)+ 2Ar(Cl)=24=2×35.5=95

MgCl₂ in excess

.n (AgCl)= n(AgNO₃) = 0.002 mol

Theoretical yield m(AgCl theor)=n×Mr=0.002× 153.5=0.307g

Mr (AgCl)=Ar (Ag) +Ar (Cl)=108 +35.5=153.5

Percent yield= $\frac{m}{m(theor)}$ ×100%= $\frac{0.292}{0.307}$ × 100%=95 %

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