Question

Consider the reaction for the preparation of the aluminum phosphate precipitate:
Na3PO4(aq)+ALCL3(aq) —> ALPO4(s)+3BaCL(aq)
If 20g of sodium phosphate react with an equal mass of aluminium chloride calculate:
A) number of moles of the precipitate formed
B)mass of remaining excess reactant
C)actual yield of the reaction if percentage yield is 80%

Answer 1

$n(Na3PO4)=\dfrac{20g}{163.94g/mol}=0.122 moles$

$n(AlCl3)=\dfrac{20g}{133.34g/mol}=0.15mol$

$n(remaining.excess.reactant(AlCl3))=0.15moles - 0.122moles=0.028moles$

$m(remaining.excess.reactant(AlCl3))=133.34g/mol*0.028moles=3.734g$

$m(AlPO4)=\dfrac{m(Na3PO4)*M(AlPO4)}{M(Na3PO4)}=\dfrac{20g*121.95g/mol}{163.94g/mol}=14.87g$

Actual yield$=14.87*80/100=11.896g$

Answer:

A)0.122moles

B)3.734g

C)11.896g