Question

Students at Gonzaga University were attempting to make barium phosphate by precipitation when mixing solutions of barium chloride and trisodium phosphate:17

BaCl2(aq) + Na3PO4(aq) â Ba3(PO4)2 + 6NaCl(aq)

In one case, 0.504 g of BaCl, and 0.611 g of Na PO, were each dissolved separately in -50 mL H,O in beakers. The Na, PO, solution was poured into the BaCl, solution with mixing and the resulting suspension of white powder was digested at 80°C for 30 min to give 0.756 g of white solid after filtration, washing, and air drying.

Identify the limiting reagent.

a. BaCI2
b. Naci
c. Na3PO4
d. Ba3(PO4)3

Answer 1

Answer:

Explanation:

3BaCl₂(aq)        +    2Na₃PO₄(aq)   =   Ba₃(PO₄)₂    +    6NaCl(aq)

3 x 208 g = 624 g    2 x 164 g = 328 g  

 328 g  of Na₃PO₄ reacts with 624 g of BaCl₂

.611 g  of Na₃PO₄ reacts with 624 x .611/328 g of BaCl₂

624 x .611/328 g = 1.16 g of BaCl₂

BaCl₂ available is .504 g which is less than required .

Hence BaCl₂ is limiting reagent .