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0.5 mol of BaCl2 is mixed with 0.2 mol...

0.5 mol of `BaCl_2` is mixed with 0.2 mol of `Na_3PO_4` . The maximum number of mol of `Ba_3(PO_4)_2` that 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
`Na_3PO_4` is limiting reagent .
`3BaCl_2 + 2Na_3 PO_4 to Ba_3 (PO_4)_2 + 6NaCl`
2 mol of `Na_3PO_4` given = 1 mole of `Ba_3(PO_4)_2` 0.2 mol of `Na_3PO_4` will give = 0.1 mol of `Ba_3(PO_4)_2`
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