What is the amount of ca3(po4)2 that can be prepared from a mixture of 9g of Ca(OH) 2 and 11g of H3po4?

1 answer:

6 0

Moles Ca(OH)2 = 12.9 /74.092 g/mol=0.174
moles H3PO4 = 18.37/98.0 g/mol=0.187 
3 : 2 = x : 0.187 
x = moles Ca(OH)2 needed =0.281 
we have only 0.174 moles of calcium hydroxide so it is the limiting reactant 
moles Ca3(PO4)2 = 0.174/3=0.0580 
mass = 0.0580 mol x 310.18 g/mol=18.0 g

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