Question

8. Calculate the maximum number of mole of Ba3(PO4)2 when 0.6 mole of BaCl2 is mixed with 0.6
mole of Na3PO4.

Answer 1

Answer on Question #41595, Chemistry, Inorganic Chemistry

Question:

8. Calculate the maximum number of mole of Ba3(PO4)2 when 0.6 mole of BaCl2 is mixed with 0.6 mole of Na3PO4.

Solution:

$$3BaCl_2 + 2Na_3PO_4 = 2Ba_3(PO_4)_2 + 6NaCl$$

According to the stoichiometric ratio, sodium phosphate is in excess; hence, barium chloride is the limiting reactant.

$$n(Ba_3(PO_4)_2) \text{ (mole)} = n(BaCl_2) \cdot 2/3 = (0.6/3) \cdot 2 = 0.4 \text{ mol}$$

Answer: 0.4 mol

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