Consider the following reaction :
3 PbCl₂+2(NH₄)₃ PO₄ → Pb₃(PO₄)₂+6 NH₄ Cl
If 72 mmol of PbCl₂ is mixed with 50 mmol of (NH₄)₃ PO₄, then the amount of Pb₃(PO₄)₂ formed is ________ mmol (nearest integer).

Solution

Chemistry ma’am · 3 steps

Divide each reactant by its coefficient: the smaller share runs out first and decides the product.

1. Find the limiting reagent. Divide each amount by its coefficient in the balanced equation. For PbCl₂: 72/3 = 24. For (NH₄)₃PO₄: 50/2 = 25.
2. Smaller share wins. 24 < 25, so PbCl₂ is the limiting reagent. It is used up completely, and some (NH₄)₃PO₄ is left over.

   n_(Pb₃(PO₄)₂) = 1/3 × 72 mmol = 24 mmol

3. So. 3 mmol of PbCl₂ give 1 mmol of Pb₃(PO₄)₂, so 72 mmol give 24 mmol.

The common slip: 25. Taking (NH₄)₃PO₄ as limiting because 50 is less than 72 gives 25, not 24. Divide by the coefficients before comparing.

The idea to keep. Never compare raw amounts of reactants. Compare amount divided by coefficient; the smallest one is limiting, and you calculate the product from that reactant only.

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