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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 |
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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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