Question

In a certain chemical reaction, 297 g of zinc chloride was produced from the single replacement reaction of excess zinc and 202.7 g of lithium chloride. What is the percent yield of zinc chloride?

Answer 1

Hey katie :

Molar mass of LiCl = 42.4 g/mol

Number of mole of LiCl = (given mass)/(molar mass)

= 202.7/42.4  => 4.78 moles of LiCl

Reaction taking place  is  :

2LiCl  +  Zn   -- >  ZnCl2   +   2Li

according to reaction :

2 mole of LiCl give 1 mole of ZnCl2

1 mole of LiCl give 1/2 mole of ZnCl2

4.78 mole of LiCl give (1/2)*4.78 mole of ZnCl2

number of mole of ZnCl2 formed = 2.39 moles

molar mass of ZnCl2 = 136.3 g/mol

mass of ZnCl2 formed = (number of moles of ZnCl2)*(molar mass)

= 2.39*136.3

= 325.8 g of ZnCl2

%yield = {(actual yiield)/(theoretical yield)}*100

= (297/325.8)*100

= 91.2 %

Answer : 91.2%

Hope that helps!