Question

A 14.585 g sample of CaCl2 was added to 12.283 g of K2CO3 and mixed in water. A 3.573 g yield of CaCO3 was obtained.

(1) What is the limiting reagent?

(2) Calculate the percent yield of CaCO3.

Answer 1

(1)

Reaction take place as

CaCl₂ + K₂CO₃   CaCO₃ + 2KCl

molar mass of CaCl₂ = 110.98 g/mol then 14.585 gm = 14.585/110.98 = 0.1314 mole

molar mass of K₂CO₃ = 138.205 g/mol then 12.283 gm = 12.283/138.205 = 0.0888 mole

According to reaction 1 mole of CaCl₂ react with 1 mole of  K₂CO₃ then for complete reaction 0.1314 mole of CaCl₂ require 0.1314 mole of K₂CO₃ but K₂CO₃ given only 0.0888 mole therefore it is limiting reagent.

Limiting reagent is K₂CO₃

(2) According to reaction 1 mole of K₂CO₃ prodece 1 mole of CaCO₃ then 0.0888 mole of K₂CO₃ prodece 0.0888 mole of CaCO₃

molar mass of CaCO₃ = 100.0869 g/mol

then 0.0888 mole = 0.0888 100.0869 = 8.8877 gm

8.8877 of CaCO₃ is theoritical yield = 100% yield then 3.573 gm =

3.573 100 /8.8877 = 40.20 %

percent yield of CaCO3 = 40.20 %