If 5 moles of BaCl₂ is mixed with 2 moles of Na₃ PO₄, the maximum number of moles of Ba₃(PO₄)₂ formed is ___________ (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Balance first, then ask which reactant runs out. Here the phosphate is gone long before the barium.

1. Balanced equation. Three Ba²⁺ ions and two PO₄³⁻ ions make one formula unit of barium phosphate.

   3BaCl₂ + 2Na₃PO₄ → Ba₃(PO₄)₂ + 6NaCl

2. Find limiting reagent. Divide each amount by its coefficient. BaCl₂: 5/3 = 1.67. Na₃PO₄: 2/2 = 1. The smaller value belongs to Na₃PO₄, so it is the limiting reagent.
3. Use limiting reagent. 2 mol Na₃PO₄ gives 2/2 × 1 = 1 mol Ba₃(PO₄)₂. This uses only 3 mol BaCl₂, so 2 mol BaCl₂ is left over.
4. So. The maximum amount of Ba₃(PO₄)₂ formed is 1 mol.

The common slip: 2. Treating BaCl₂ as limiting gives 5/3 ≈ 1.67, which rounds to 2. There is not enough phosphate to make that much.

The idea to keep. Divide each reactant's moles by its coefficient in the balanced equation. The smallest value marks the limiting reagent, and the product is calculated from that reactant alone.

Answer from the NTA final answer key. Solution by PrepWiser.