How much `Na_(2)SO_(4)` will be required for complete precipitation of 7.336g `BaCl_(2)` dissolved in water? What will be the mass of `BaSO_(4)` precipitated? [Ba=137.36]

Explanation

`underset((137.36+2xx35.5)=208.36g)(BaCl_(2))+underset((2xx23+32+64)g=142g)(Na_(2)SO_(4)) to underset((137.36+32+64)=233.36g)(BaSO_(4))+2NaCl`
Now, 208.36g `BaCl_(2)-=142gNa_(2)SO_(4)-=233.36gBaSO_(4)`
`therefore 7.336gBaCl_(2)-=5gNa_(2)SO_(4)-=8.216gBaSO_(4)`
`therefore`5g of `Na_(2)SO_(4)` will be required for complete precipitation of 7.336g `BaCl_(2)` and the mass of `BaSO_(4)` precipitate is 8.216g.