For the reactions shown below, we added 5.00 mL of 0.0370 M NaOH to a test tube
ID: 568007 • Letter: F
Question
For the reactions shown below, we added 5.00 mL of 0.0370 M NaOH to a test tube containing one of the two cations (Ni2+ or Fe3+) and recovered 0.00659 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.)
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+?
Explanation / Answer
nO of mol of NnaOH used = M*V
= 0.037*5
= 0.185 mol
1 mol Ni(OH)2(s) = 2 mol NaOH(aq)
No of mol of Ni(OH)2(s) = 0.185/2 = 0.0925 mol
Amount of Ni(OH)2(s) = 0.0925*92.708 = 8.57 g
1 mol Fe(OH)3(s) = 3 mol NaOH(aq)
No of mol of Fe(OH)3(s) = 0.185/3 = 0.06167 mol
Amount of Fe(OH)3(s) = = 0.06167*88.86 = 5.5 g
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.