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Molodets [167]
3 years ago
9

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?

Chemistry
1 answer:
schepotkina [342]3 years ago
6 0
Moles Ca(OH)2 = 12.9 /74.092 g/mol=0.174 
<span>moles H3PO4 = 18.37/98.0 g/mol=0.187 </span>
<span>3 : 2 = x : 0.187 </span>
<span>x = moles Ca(OH)2 needed =0.281 </span>
<span>we have only 0.174 moles of calcium hydroxide so it is the limiting reactant </span>
<span>moles Ca3(PO4)2 = 0.174/3=0.0580 </span>
<span>mass = 0.0580 mol x 310.18 g/mol=18.0 g</span>
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