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For the reactions shown below, we added $\pu{5.00 mL}$ of $\pu{0.0390 M}$ $\ce{NaOH}$ to a test tube containing one of the two cations $\ce{Ni^2+}$ or $\ce{Fe^3+}$ and recovered $\pu{0.00695 g}$ of precipitate. \begin{align} \ce{Ni(NO3)2(aq) + 2 NaOH(aq) &-> Ni(OH)2(s) + 2 NaNO3(aq)}\\ \ce{Fe(NO3)3(aq) + 3 NaOH(aq) &-> Fe(OH)3(s) + 3 NaNO3(aq)} \end{align}

  1. How much precipitate in moles would be recovered theoretically if the ion was $\ce{Ni^2+}$?

For this answer I got $\pu{0.0000975 mol}$, but that is wrong and I don't understand why.

  1. How much precipitate in moles would be recovered theoretically if the ion was $\ce{Fe^3+}$?

  2. How much precipitate in grams would be recovered theoretically if the ion was $\ce{Ni^2+}$?

  3. How much precipitate in grams would be recovered theoretically if the ion was $\ce{Fe^3+}$?

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As far as I can see your calculation checks out. Maybe it's an issue with the answer checking system?

For the amount of substance of precipitate if $\ce{Fe^3+}$ is the ion, you carry out a very similar calculation as before. You find the amount of substance of $\ce{NaOH}$ in the $\pu{5 mL}$ solution and then divide that by 3 to find the amount of substance of $\ce{Fe(NO3)3}$ in this particular reaction. Pay attention to the relative reacting amounts of reactants. The product has the same amount of substance as $\ce{Fe(NO3)3}$ so that's your answer for question 2.

questions 3 and 4 are very simple. Now that you have the theoretical amount of substance of the precipitates you can convert them into masses by manipulating the \begin{align} \frac{\text{formula mass}}{ \text{relative molecular mass}} &= \text{amount of substance}\\ \Longleftrightarrow \text{amount of substance} \times \text{relative molecular mass} &= \text{mass} \end{align}

You can get the relative molecular masses of $\ce{Ni(OH)2}$ and $\ce{Fe(OH)3}$ by using your periodic table.

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