1.

0.5 mol of BaCl_2is mixed with 0.2 mol ofNa_3PO_4 . The maximum number of mol of Ba_3(PO_4)_2that can be formed is :

Answer»

0.7
0.5
0.3
0.1

Solution :`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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