How many grams of Ca3(PO4)2 precipitate can form by reacting 169.1 mL of 1.9 M C

Question

How many grams of Ca3(PO4)2 precipitate can form by reacting 169.1 mL of 1.9 M CaBr2 with an excess amount of Li3PO4, given the balanced equation:
2 Li3PO4 (aq) + 3 CaBr2 (aq) --> 6 LiBr2 (aq) + Ca3(PO4)2 (s) How many grams of Ca3(PO4)2 precipitate can form by reacting 169.1 mL of 1.9 M CaBr2 with an excess amount of Li3PO4, given the balanced equation:
2 Li3PO4 (aq) + 3 CaBr2 (aq) --> 6 LiBr2 (aq) + Ca3(PO4)2 (s) How many grams of Ca3(PO4)2 precipitate can form by reacting 169.1 mL of 1.9 M CaBr2 with an excess amount of Li3PO4, given the balanced equation:
2 Li3PO4 (aq) + 3 CaBr2 (aq) --> 6 LiBr2 (aq) + Ca3(PO4)2 (s)

Explanation / Answer

according to the given data

no.of moles of CaBr2= molarity*volume/1000 = 1.9*169.1/1000 =0.32129 moles of CaBr2

according to the reaction

3 moles of CaBr2 on reCTION WITH LITHIUMPHOSPHATE GIVES----------------1 MOLE OF CALICIUMPHOSPHATE

0.32129 MOLES OF CaBr2 react with lithium phospahte gives --------------------------------------?

                       = 0.32129/3

                       =0.107 moles

mass of calicium phospahate is formed = noof moles *molar mass = 0.107*310.176 = 33.18 grams of cliciumphosphate

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