If 0.50 mol of `BaCl_(2)` is mixed with 0.20 mol of `Na_(3)PO_(4)`, the maximum number of moles of `Ba_(3)(PO_(4))_(2)` that can be formed is

Options

A. `0.7`
B. `0.5`
C. `0.3`
D. `0.1`

Answer

D

Explanation

`2Na_(3)PO_(4)+3BaCl_(2)toBa_(3)(PO_(4))_(2)darr+6NaCl`
`3molBaCl_(2)=2mol` of `Na_(3)PO_(4)`
0.5 mol `BaCl_(2)=(2)/(3)xx0.5` mol of `Na_(3)PO_(4)`
`=0.33 mol` of `Na_(3)PO_(4)`
But mole of `Na_(3)PO_(4) = 0.2`
`:. Na_(3)PO_(4)` is the limiting reactant.
2 mole of `Na_(3)PO_(4)` gives 1 mole of `Ba_(3)(PO_(4))_(2)`
`:.` Max. `Ba_(3)(PO_(4))_(2)` formed = 0.1 mol`