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3 years ago

How many grams of calcium phosphate can be produced when 89.3 grams of calcium chloride reacts with excess sodium phosphate?

1 answer:

4 0

Molecular weights:
CaCl2 = 40 + 35.5*2 = 111 g/mol
Ca3(PO4)2 = 40*3 + (31+16*4)*2 = 310 g/mol
89.3 g CaCl2 = 89.3/111 = 0.8045 mol CaCl2

Balanced equation:
3CaCl2 + 2Na3PO4 --------> 6NaCl + Ca3(PO4)2

3 moles CaCl2 produce 1 mol Ca3(PO4)2
Therefore 0.8045 mol CaCl2 produces 1/3 * 0.8045
= 0.2682 mol Ca3(PO4)2
= 0.2682 mol * 310 g/mol
= 83.1 g Ca3(PO4)2

Ans: 83.1 g

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For overall reaction on adding a and b we get c

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The equilibrium constant for the overall reaction is $K_a\times K_b$

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