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(Solved): For The Reactions Shown Below, We Added 4.50 ML Of 0.0380 M NaOH To A Test Tube Containing One Of Th ...

For the reactions shown below, we added 4.50 mL of 0.0380 M NaOH to a test tube containing one of the two cations (Ni2+ or Fe

For the reactions shown below, we added 4.50 mL of 0.0380 M NaOH to a test tube containing one of the two cations (Ni2+ or Fe3+) and recovered 0.00793 g of precipitate. Ni(NO3)2(aq) + 2NaOH(aq) ? Ni(OH),(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.) 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+? 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+ O Fe3+

Expert Answer


1) Number of moles = mass / molar mass Number of moles of Ni(OH)2 would be recovered = 0.00793g/92.709g/mol = 0.00008554 mol 2) Number of moles of Fe(OH)3 would
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