500mL of 0.25M `Na_(2)SO_(4)` solution is added to an aquesous solution is 15g of `BaCl_(2)` resulting in the formation of a white precipatate of insoluble `BaSO_(4)`. How many moles and how many grams of `BaSO_(4)` are formed.

Explanation

`500cm^(3)` of `0.25M Na_(2)SO_(4)` solution contain `Na_(2)SO_(4)=(0.25)/(1000)xx500=0.125" mole"`
`15g BaCl_(2)=(15)/(208)" mole"=0.072 " mole"`
`Na_(2)SO_(4)+BaCl_(2) rarr BaSO_(4)+2NaCl`
Evidently, `BaCl_(2)` will be the limiting reactant.
`BaSO_(4)` formed = 0.072 mole = `0.072xx233g=16.776g`