If 0.5 mole of `BaCl_(2)` is mixed with 0.20 mole of `Na_(3)PO_(4)`, the maximum number of `Ba_(3)(PO_(4))_(2)` that can be formed is :

Options

A. 0.7mol
B. 0.5mol
C. 0.2mol
D. 0.1mol

Answer

D

Explanation

Let us first solve this Problem by writing the complete balanced reaction.
`3BaCI_(2)+2Na_(3)PO_(4)toBa_(3)(PO_(4))darr+6NaCI`
We can see that the moles of `BaCI_(2)` required would be `0.2 xx (3)/(2) = 0.3`. Since `BaCI_(2)` `BaCI_(2)` used are `(3)/(2)` times the moles of `Na_(3)PO_(4)`. Therefore, to react with 0.2 mol of `NA_(3)PO_(4)`, the moles of `BaCI_(2)` required would be `0.2xx(3)/(2)=0.3`. Since `BaCI_(2)` is 0.5 mol, we can conclude that `Na_(3)PO_(4)` is the limiting reagent. Therefore, moles of `Ba_(3)(PO_(4))_(2)` formed is `0.2xx(1)/(2)=0.1 mol`.