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If 0.5 mole of BaCl(2) mixed with 0.20 m...

If 0.5 mole of `BaCl_(2)` mixed with 0.20 mole of `Na_(3) PO_(4)` the maximum number of moles of `Ba_(3)(PO_4)_(2)` then can be formed is

A

0.7

B

0.5

C

0.3

D

0.1

Text Solution

Verified by Experts

The correct Answer is:
D
`3BaCl_(2) + 2Na_(3)PO_(4) rarr Ba_(3)(PO_(4))_(2) + 6NaCl`
Limiting reactant is `Na_(3)PO_(4)`., 0.2 mol of `Na_(3)PO_(4)` will give `Ba_(3)(PO_(4))_(2) = (1)/(2) xx 0.2 = 0.1` mol of `Ba_(3)(PO_(4))_(2)`
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