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?

Question

Yuliana Solomon

Chemistry

12 January, 00:40

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?

+4

Answers (1)

Know the Answer?

Elyse Bennett · 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