Question

Use the following reaction:

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

How much barium chloride is produced (in g) when 1500 g of sodium chloride react with excess barium phosphate?

Round your answer to a whole number, do not write the units.

Answer 1

Mass of Barium chloride produced : 2666.56 g

Further explanation

Given

Reaction

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

1500 g of sodium chloride

Required

Barium chloride produced

Solution

mol NaCl (MW= 58.5 g/mol) :

mol = mass : MW

mol = 1500 g : 58.5 g/mol

mol = 25.64

mol Barium chloride-BaCl2 based on mol NaCl(as a limiting reactant)

From the equation, mol ratio NaCl : BaCl2 = 6 : 3, so mol BaCl2 :

`\tt (3)/(6)* 25.64=12.82`

mass BaCl2(MW=208 g/mol) :

mass = mol x MW

mass = 12.82 x 208

mass = 2666.56 g