Question

For the reactions shown below, we added 5.00 mL of 0.0340 M NaOH to a test tube containing one of the two cations (Ni2+ or Fe3+)and recovered 0.00788 g of precipitate. Ni(NO3)2(aq) + 2 NaOH(aq) Ni(OH)2(s) + 2 NaNO3(aq) Fe(NO3)3(aq)+ 3 NaOH(aq)-Fe(OH)3(s)3 NaNO3(aq) How much precipitate in moles would be recovered theoretically if the ion was Ni2+? (Enter an unrounded value. Use at least one more digit than given.) X mol How much precipitate in moles would be recovered theoretically if the ion was Fe3+? (Enter an unrounded value. Use at least one more digit than given.) mol How much precipitate in grams would be recovered theoretically if the ion was Ni2+? 402 How much precipitate in grams would be recovered theoretically if the ion was Fe3+? Based on the precipitate amount recovered, which of the two ions was in the unknown? Ni2+ Fe3+

Answer 1

Answer)

I have enclosed answers in rectangular boxes for your convenience.

Volume of NaoH = 5 . 00 mL concentration of NaoH = 0 . 034 oM so moles of NaoH = 5 . 00 mL times 0 . 034 moles/1000mL 1L = 1000 mL = 1 . 7 times 106-4 moles Reactions given Ni(Mo_3) (g) + 2MaoH (9q) rightarrow Ni(oH)_2(s) + 2MaNo_3 (9q) Fe(No_3)_3(aq) + 3NaoH(9q) rightarrow Fe (OH)_3(s) + 3Na(Mo_3) (9q) \r\n Rightarrow first we have to find out ratio of mole of NaoH to ratio of preciptitates in lach reaction: Rightarrow In reaction Ni(No_3)_2(9q) + 2 NaoH(9q) rightarrow Ni(OH)_2(s) + 2Na(No_3(9q) 1 mole of Ni(No_3)_2 (9q) reacts with 2 moles of NaoH and produces the 1 mole of Ni(OH)_2(s) precipitate and 2 mole of NaNo_3(9q) so ratio of NaoH moles to Ni(OH)_2 (s) moles Rightarrow 2: 1 Note: - we are taking Ni(OH)_2 (s) as precipitates not NaNo_3(9q) because only Ni(OH)_2(s) precipitates out but NanO_3 (9q) remains in dissolved form in solution Rightarrow In reaction Fe(No_3)_3(q) + 3 NaoH (9q)
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