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# Calculate the amounts of the substance in order to prepare a certain amount of the solution

wrote...
Posts: 9
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5 years ago
 Calculate the amounts of the substance in order to prepare a certain amount of the solution    I have tried using Molarity= Moles of Solute/Liters of Solution, but i cannot figure the answer out. If you can only get the first two questions, that is okay, thank you!1) Calculate the amount of potassium ferrocyanide trihydrate [K4Fe(CN)6•3H2O] needed to prepare 250 ml of 0.025 M solution.  2) Calculate the amount of zinc sulfate heptahydrate [ZnSO4•7H2O] needed to prepare 250 ml of 0.050 M solution.  3) Assuming that the actual molarities of your stock solutions are equal to the target molarities given above, calculate what titration volume of zinc sulfate would be expected for titrating  a 25.00 ml aliquot of the ferrocyanide solution for each of the five  possible reactions presented below:    K4Fe(CN)6(aq)  +  ZnSO4(aq) -----> K2ZnFe(CN)6(s)  +  K2SO4(aq)   K4Fe(CN)6(aq)  +  2 ZnSO4(aq) ----> Zn2Fe(CN)6(s)  +  2 K2SO4(aq)   2 K4Fe(CN)6(aq)  +  ZnSO4(aq) ----> K6Zn[Fe(CN)6]2(s)  +  K2SO4(aq)   2 K4Fe(CN)6(aq)  +  3 ZnSO4(aq) ----> K2Zn3[Fe(CN)6]2(s)  +  3 K2SO4(aq)   3 K4Fe(CN)6(aq)  +  2 ZnSO4(aq) ----> K8Zn2[Fe(CN)6]3(s)  +  2 K2SO4(aq) Read 917 times 11 Replies

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wrote...
Educator
5 years ago
 For #1 and #2, see if this helps before I continue Attached file Thumbnail(s): You must login or register to gain access to this attachment.
wrote...
5 years ago
 Where did the 3 come from on the outside of the bracket in problem 1?
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Educator
5 years ago
 I thought it contained it... opps.
wrote...
5 years ago
 The MWt of [K4Fe(CN)6•3H2O] (C6H6FeK3N6O3) = 383.3 g 250 mL of 0.025 M K4Fe(CN)6•3H2O contains (250/1000)×0.025 = 6.25×10^-3 mole Hence amount needed = (6.25×10^-3)×383.3 = 2.40 g (should only be 2 sig figs) Source https://biology-forums.com/index.php?action=downloads;sa=view;down=14860
wrote...
5 years ago
 I have tried using Molarity= Moles of Solute/Liters of Solution, but i cannot figure the answer out. I have the first two, but the third question is confusing.
wrote...
Educator
5 years ago
 I'm unsure how to tackle it too
wrote...
3 years ago
 Thank you
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2 months ago
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A month ago
 thank you
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A month ago
 3. Molarity of ZnSO4. 7H2O = 0.050 MVolume = 250 mL = 0.250 LMoles of ZnSO4. 7H2O = Molarity * Volume = 0.050 M * 0.250 L = 0.0125 molesMolar mass of ZnSO4. 7H2O = 287.5496 g/molMass of  ZnSO4. 7H2O = Moles * Molar mass = 0.0125 moles * 287.5496 g/mol = 3.59 g
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A month ago
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