A reaction is carried out by combining 65.00 mL of a 0.108 M sample of (NH4)3PO4...

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Chemistry

A reaction is carried out by combining 65.00 mL of a 0.108 Msample of (NH4)3PO4 and 48.00 mL of a 0.230 M Fe9NO3)2 solution.The net ionic equation is :

3Fe^2+(aq) + 2PO4^3-(aq) ---> Fe3(PO4)2(s) balanced

1a) What is the mass of iron(II) phosphate (MM= 357.49 g/mol)produced in the reaction?

1b) What are the concentrations of the reacting ions (Fe^2+(aq)and PO4^3+(aq)) at the end of the reaction (assuming no change involume)?

Answer & Explanation Solved by verified expert
3.8 Ratings (447 Votes)
3 FeNO32 2 NH43PO4 Fe3PO42s 6 NH4NO3 i Complete Ionic Reaction is written as 3Fe2 6NO3 6NH4 2PO43 Fe3PO42s 6NO3 6NH4ii And Net Ionic Equation is 3Fe2aq 2PO43aq Fe3PO42siii Now given that NH43PO4 0108 and V 65 ml Moles of NH43PO4 0108 molL0065 L 000702 mol FeNO32 0230 M    See Answer
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