1.

If 0.5 mole of `BaCl_(2)` are mixed with 0.2 mole of `Na_(3)PO_(4)`,the maximum number of moles. Of `Ba_(3)(PO_(4)_(2))` that can be formed, isA. 0.7B. 0.5C. 0.3D. 0.1

Answer» d) `3BaCl_(2) + 2Na_(3)PO_(4) to Ba_(3)(PO_(4))_(2) + 3NaCl`
Limiting reactant is `Na_(3)PO_(4)`.
0.2 mole of `Na_(3)PO_(4)` will give = `1/2 xx 0.2`
=0.1 mole of `Ba_(3)(PO_(4))_(2)`


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