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If `0.50 mol` of barium chloride `(BaCl_(2))` is mixed with `0.20` mol of sodium phosphate `(Na_(3) PO_(4))` the maximum number of moles of barium phosphate `[Ba_(3)(PO_(4))_(2)]` formed isA. `0.25`B. `0.10`C. `0.40`D. `0.50` |
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Answer» Correct Answer - B `underset(3 mol)(3BaCl_(2)) + underset(2 mol) (2Na_(3) PO_(4)) rarr underset(1 mol) (Ba_(3)(PO_(4))_(2)) + underset(6 mol) (6NaCl)` `0.50 mol BaCl_(2) xx (1 mol Ba_(3) (PO_(4))_(2))/(3 mol BaCl_(2)) = 0.17 mol Ba_(3) (PO_(4))_(2)` `0.20 mol Na_(3) PO_(4) xx (1 mol Ba_(3) (PO_(4))_(2))/(2.0 mol Na_(3) PO_(4)) = 0.1 mol Ba_(3) (PO_(4))_(2)` Therefore, `Na_(3) PO_(4)` is the limiting rectant and the maximum moles of `Ba_(3) (PO_(4))_(2)` formed is `0.1` mol. |
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