`10 mL` of `1 M BaCl_(2)` solution and `5 mL 0.5 N K_(2)SO_(4)` are mixed together ot 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

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