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. `0.70 mol`
B. `0.50 mol`
C. `0.20 mol`
D. `0.10 mol`

Answer

D

Explanation

(i) Write balanced chemical equation for chemical change.
(ii) Find limiting reagent.
(iii) Amount of product formed formed will be determined by amount of limiting reagent.
The balanced equation is:
`underset(No . of "moles")()" "underset(0.5 3 "moles")(3BaCl_(2))+underset(0.2 "moles")(2Na_(3)PO_(4))rarrunderset(1 "mole")(Ba_(3)(PO_(4))_(2)+6"Nacl")`
Limiting reagent is `Na_(3)PO_(4)(0.2 mol), BaCl_(2)` is in excess. From the above equation:
`2.0` moles of `Na_(3)PO_(4)` yields `Ba_(3)(PO_(4))^(2)= 1` mole
`:. 0.2` moles of `Na_(3)PO_(4)` will yield `Ba_(3)(PO_(4))_(2)`
`=1/2xx0.2=0.1` mol.