Question

0.7 moles of potassium sulphate is allowed to react with 0.9 moles of barium chloride in aqueous solution. The number of moles of the substance precipitated in the reaction is:
A. 1.4 moles of potassium chloride
B. 0.7 moles of barium sulphate
C. 1.6 moles of potassium chloride
D. 1.6 moles of barium sulphate

Answer 1

Hint: A precipitate is formed when salts of halogens react with acids containing sulphate. The precipitate formed is white in color. A limiting reagent is the substance present in less amount.

Complete step by step solution:
We have been given 0.7 moles of potassium sulphate,${{K}_{2}}S{{O}_{4}}$ , that reacts with 0.9 moles of barium chloride,$BaC{{l}_{2}}$ in an aqueous medium. We have to find the moles of the precipitate formed. For this the reaction between these compounds is considered, and written as,
\[{{K}_{2}}S{{O}_{4}}+BaC{{l}_{2}}\to BaS{{O}_{4}} \downarrow +2KCl\]
This reaction clearly shows that, when 1 mole of potassium sulphate is taken, alongside, 1 mole of barium chloride, the product or the precipitate formed is 1 mole of barium sulphate, and 2 moles of potassium chloride.
So, through stoichiometric , or unitary method we can find the moles of barium sulphate to be the same as potassium sulphate and barium chloride, which is 0.7 moles.
Moles of $BaS{{O}_{4}}$in 0.7 moles of reactants = $\dfrac{1\,mole\,BaS{{O}_{4}}}{1\,mole\,{{K}_{2}}S{{O}_{4}}}\times 0.7mol\,{{K}_{2}}S{{O}_{4}}$
Moles of $BaS{{O}_{4}}$in 0.7 moles of reactants = 0.7 moles of $BaS{{O}_{4}}$
Hence, 0.7 moles of barium sulphate are precipitated in the product.

Note: SInce moles of barium chloride are 0.9 moles, and potassium sulphate are 0.7 moles, but only 0.7 moles of barium chloride will react to produce 0.7 moles of precipitate, barium sulphate. This shows that sulphate ions are the limiting reagents in this process.