`10 mL` of `1 M BaCl_(2)` solution & `5 mL 0.5 M K_(2)SO_(4)` are mixed together to precipitate out `BaSO_(4)`. The amount of `BaSO_(4)` precipated will be

Options

A. 0.005 mol
B. 0.00025 mol
C. 0.025 mol
D. 0.0025 mol

Answer

D

Explanation

`{:(BaCl_(2),+,K_(2)SO_(4),rarr,BaSO_(4)darr,+,2KCl),(10 xx 1,,5 xx 0.5,,,,),(="10 m mol",,= "2.5 m mol",,,,):}`
Here, `K_(2)SO_(4)` is the limiting reagent, 1 mmol `K_(2)SO_(4) -= 1` mmol `BaSO_(4)`
2.5 m mol `K_(2)SO_(4) -= 2.5 m` mol `BaSO_(4) = (2.5)/(1000) = 0.0025` mol `BaSO_(4)`