1.

500cm^(3) of 0.250 M Na_(2)SO_(4) solution added to an aqueous solution of 15.00 g of BaCl_(2) resulted in the formation of a white precipitate of BaSO_(4). How many moles and how many grams of BaSO_(4) are formed ?

Answer»

SOLUTION :`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`


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