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

Answer

D

Explanation

Let us first solve this Prob by writing the complete balanced reaction.
`3BaCl_(2) + 2 Na_(3)PO_(4) to Ba_(3)(PO_(4))_(2) downarrow + 6NaCl`
We can see that the moles of `BaCl_(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 `BaCl_(2)` required would be `0.2 xx 3/2 = 0.3` . Since `BaCl_(2)" is " 0.5 ` mol, wer can conclude that `Na_(3)PO_(4)` is the limiting reagent. Therefore, moles of `Ba_(3)(PO_(4))_(2)` formed is `0.2 xx 1/2 = 0.1 ` mol .