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If 0.50 mole of BaCI2 is mixed with 0.20 mole of Na3PO4, the maximum number of moles of Ba3 (PO4)2 that can be formed is a. 0.70 b. 0.50 c. 0.20 d. 0.10

If 0.50 mole of BaCI2 is mixed with 0.20 mole of Na3PO4, the maximum number of moles of Ba3 (PO4)2 that can be formed is
a. 0.70
b. 0.50
c. 0.20
d. 0.10

Grade:upto college level

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 years ago
Sol. The balanced chemical reaction is 3 BaCI2 + 2Na3PO4→ Ba3 (PO4)2 + 6NaCI In this reaction, 3 moles of BaCI2 combines with 2 moles of Na3 PO4. Hence, 0.5 mole of ofBaCI2 require 2/3× 0.5 = 0.33 mole of Na3PO4. Since available Na3PO4 (0.2 mole) is less than required mole (0.33), it is the limiting reactant and would determine the amount of product Ba3 (PO4)2. ∵ 2 moles of Na 3 PO4 gives 1 mole Ba3 (PO4)2 ∴0.2 mole of Na 3PO4 would give 1/2 × 0.2 = 0.1 mole Ba 3(PO4)2

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