Fe, (aq) + 2NaOH(aq) Fe(OH)2 (s) + 2Na. (aa) he cation Na\' is a spectator ion a
ID: 589606 • Letter: F
Question
Fe, (aq) + 2NaOH(aq) Fe(OH)2 (s) + 2Na. (aa) he cation Na' is a spectator ion and is excluded from the net ionic equation. The balanced chemical equation can also be used to develop stoichiometric ratios his reactions shows that 2 moles of OH are needed to precipitate each mole of Fe (aq). Part D Let us assume that Fe(OH)2 (s) is completely insoluble, which signifies that the precipitation reaction with NaOH(a) (presented in the transition) would go to completion Fe2. (aq) + 2NaOH(aq) Fe(OH)2 (s) + 2Na+ (aq) If you had a 0.350 L solution containing 0 0240 Mof Fe? (aq), and you wished to add enough 1.39 M NaOH(aq) to precipitate minimum amount of the NaOH (aq) solution you would need to add? Assume that the NaOH(o) solution is the only source of OH (aq) for the precipitation Express the volume to three significant figures and include the appropriate units Hints all of the metal, what is the ? minimal volume of NaOH(aq) solution 11.2 ml My Answers Give Up Incorrect: Try Again; 5 attempts remaining The required volume of NaOH solution can be calculated using the inital volume olee Fet' (aq) soluton the concentration of Foc (oq) recal that the units molarity are equivalent to mol/L) the coe cients from the balanced chemical equation, and the concer rason of a H De Plop as ton map determine how to apply these conversion factors, and make certain to include units when you set up your calculations to verify that undesired units cancel You can determine the volume of NaOH solution needed in units of iters
Explanation / Answer
Fe^2+ (aq) + 2NaOH(aq) ------------> Fe(OH)2(s) + 2Na^+ (aq)
1 mole 2 moles
Fe^2+ NaOH
M1 = 0.024M M2 = 1.39M
V1 = 0.35L = 350ml V2 =
n1 = 1 n2 = 2
M1V1/n1 = M2V2/n2
V2 = M1V1n2/M2n1
= 0.024*350*2/1.39*1 = 12.08ml
volume of NaOH = 12.08ml >>>>answer
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