`3 BaCl_(2) + 2Na_(3)PO_(4) to Ba_(3)(PO_(4))_(2)+6NaCl`
Maximum amount of `Ba_(3)(PO_(4))_(2)` formed when 2 moles each of `Na_(3)PO_(4)` and `BaCl_(2)` react is

Options

A. 4 mol
B. 1 mol
C. `(2)/(3) mol`
D. `(1)/(3) mol`

Answer

C

Explanation

`3 BaCl_(3)+2Na_(3)PO_(4)to Ba_(3)(PO_(4))_(2)+6NaCl`
Here `BaCl_(2)` is the limiting reactant
No. of moles of `Ba_(3)(PO_(4))_(2)` formed
`=(1)/(3)xx"N0. of moles of" BaCl_(2) = (2)/(3)` mol