Question

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?

Answer 1

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