katie112102
Answered

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 :

Willchemistry

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!

Eduard22sly

The percentage yield of zinc chloride, ZnCl₂ obtained from the reaction is 91.6%

Balanced equation

Zn + 2LiCl —> ZnCl₂ + 2Li

Molar mass of LiCl = 7 + 35.5 = 42.5 g/mol

Mass of LiCl from the balanced equation = 2 × 42.5 = 85 g

Molar mass of ZnCl₂ = 65 + (35.5×2) = 136 g/mol

Mass of ZnCl₂ from the balanced equation = 1 × 136 = 136 g

SUMMARY

From the balanced equation above,

85 g of LiCl reacted to produce 136 g of ZnCl₂

How to determine the theoretical yield of ZnCl₂

From the balanced equation above,

85 g of LiCl reacted to produce 136 g of ZnCl₂

Therefore,

202.7 g of LiCl will react to produce = (202.7 × 136) / 85 = 324.32 g of ZnCl₂

How to determine the percentage yield

  • Actual yield of ZnCl₂ = 297 g
  • Theoretical yield of ZnCl₂ = 324.32 g
  • Percentage yield =?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (297 / 324.32) ×100

Percentage yield = 91.6%

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