# How to distinguish between iron and nickel based on the mass of precipitate formed with sodium hydroxide?

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 $$\ce{Fe^3+}$$?

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.