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

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 .

it's c. elements are placed in rows based on how may valencies electrons have.

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.