If 5.05 g of ca(oh)2 and 6.70 g of h3po4 are available, how many moles of ca3(po4)2 will you produce from each?

a) 0.023 mol Ca3(PO4)2 from 5.05 g Ca(OH)2

b) 0.034 mol Ca3(PO4)2 from 6.70 g H3PO4

Explanation:

∴ Mw Ca(OH)2 = 74.093 g/mol

∴ Mw H3PO4 = 97.994 g/mol

a) n Ca3(PO4)2 = (5.05 g Ca(OH)2)×(mol Ca(OH)2/74.093 g)×(mol Ca3(PO4)2/3 mol Ca(OH)2) = 0.023 mol Ca3(PO4)2

b) n Ca3(PO4)2 = (6.70 g H3PO4)×(mol H3PO4/97.994 g)×(mol Ca3(PO4)2/2 mol H3PO4) = 0.034 mol Ca3(PO4)2

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answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error