Question

500 mL of 0.250 M `Na_(2)SO_(4)` solution is treated with 15.00 g of `BaCl_(2)`. Moles of `BaSO_(4)` formed are
A. `0.72`
B. `0.072`
C. `0.168`
D. `0.0168`

Answer 1

Correct Answer - B
`Na_(2)SO_(4)+BaCl_(2)toBaSO_(4)+2NaCl`
Molar mass of `BaCl_(2) = 137.2+2xx35.5`
`= 208.4 g mol^(-1)`
Moles of `BaCl_(2) = (15)/(208.4g mol^(-1)) = 0.072 mol`
Moles of `Na_(2)SO_(4) = ((500)/(1000)L)xx(0.250 mol L^(-1))`
`= 0.125 mol`
Here `BaCl_(2)` is the limiting reactant
`:.` Moles of `BaSO_(4)` formed = Moles of `BaCl_(2)`
`= 0.072 mol`