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.

Answer

A

Explanation

`BaCl_(2)(aq) + Na_(2)SO_(4)(aq) to BaSO_(4)(s) + 2NaCl`
1000 mL of 0.250 M `Na_(2)SO_(4)` contain `Na_(2)SO_(4)`
= 0.250 mol
500 mL of 0.250 M `Na_(2)SO_(4)` contain `Na_(2)SO_(4)`
`=0.250/1000 xx 500 = 0.125` mol
Moles of `BaCl_(2) = 15/208 = 0.0721` mol
`therefore BaCl_(2)` is limiting reactant.
Molecular mass of `BaSO_(4) = 233`
1 mol of `BaCl_(2)` reacts with `Na_(2)SO_(4)` to give
= 233 g `BaSO_(4)`
0.072 mol of `BaCl_(2)` will react with `Na_(2)SO_(4)` to gives
`= 233 xx 0.072 = 16.78`g