# Ion molar 2

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• ﻿﻿4) Calculate the moles of each ion present and convert to Molarity with the new volume. - Fe2+: Ø M ﻿since it was an ion in the ppt and the limiting react.  - PO43-: .595 moles Fe(PO4﻿)2 - .46 moles Fe(PO4﻿)2 = .135 moles .135 moles Fe(PO4﻿)2(2 moles PO43-/1 mol  Fe(PO4﻿)2) =.27 moles  .27 moles PO43- / .500 liters total = .54  M  PO43-
• ﻿﻿3) Calculate the grams of precipitate product with the moles of each reactant ﻿to figure out the limiting reactant.  3) Fe(NO3)2 :  1.38 moles ( 1 mole Fe3(PO4)2 / 3 mole Fe(NO3)2= .46 moles Fe3(PO4)2      K3PO4 : 1.19 moles  ( 1 mole Fe3(PO4)2 / 2 mole K3PO4 ) = .595 moles Fe3(PO4)2 Therefore the limiting reactant is Fe(NO3)2
• Ion molarity problem cont ...﻿
• cont on next pg ...
• ﻿Answer:  - K+ = 7.14 M  - PO43- = .27 M  - Fe2+ = Ø M - NO3- = 5.52 M
• ﻿﻿4) Calculate the moles of each ion present and convert to Molarity with the new volume continue... - NO3-:  (1.38 mol Fe(NO3)2) (2 mol NO3-/ 1 mol Fe(NO3) = 2.76 mol NO3-   2.76 mol NO3-  / .500 liters = 5.52 M  NO3-   -   K+ :( 1.19 mol K3PO4) (3 mol K+ / 1 mol K3PO4) = 3.57 mol K+  3.57 mol K+  / .500 Liters = 7.14 M   K+
• ALL DONE!!!!﻿
• ﻿2 DAYS LATER ...
• Thank you!
• Good Job Maria!
• ﻿A+
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