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If 0.50 mol of `BaCl_(2)` is mixed with 0.20 mol of `Na_(3)PO_(4)`, the maximum number of moles of `Ba_(3)(PO_(4))_(2)` that can be formed isA. (a)0.70B. (b)0.50C. (c )0.20D. (d)0.10 |
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Answer» Correct Answer - D (i) Write balanced chemical equation for chemical change. (ii) Find limiting reagent. (iii) Amount of product formed formed will be determined by amount of limiting reagent. The balanced equation is: `underset(No . of "moles")()" "underset(0.5 3 "moles")(3BaCl_(2))+underset(0.2 "moles")(2Na_(3)PO_(4))rarrunderset(1 "mole")(Ba_(3)(PO_(4))_(2)+6"Nacl")` Limiting reagent is `Na_(3)PO_(4)(0.2 mol), BaCl_(2)` is in excess. From the above equation: `2.0` moles of `Na_(3)PO_(4)` yields `Ba_(3)(PO_(4))^(2)= 1` mole `:. 0.2` moles of `Na_(3)PO_(4)` will yield `Ba_(3)(PO_(4))_(2)` `=1/2xx0.2=0.1` mol. |
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