How many grams of iron(III) hydroxide (106.87 g/mol) will precipitate if excess iron(III) nitrate is added to 0.850 L of 1.26 M sodium hydroxide?

Question

Quizplus · Accepted Answer

The reaction between iron(III) nitrate and sodium hydroxide can be represented by the balanced chemical equation: Fe(NO3)3 + 3NaOH -> Fe(OH)3 + 3NaNO3. From the equation, 1 mole of Fe(NO3)3 reacts with 3 moles of NaOH to produce 1 mole of Fe(OH)3. Given that the concentration of NaOH is 1.26 M in 0.850 L, the moles of NaOH are 1.26 mol/L * 0.850 L = 1.071 moles. Since it takes 3 moles of NaOH to react with 1 mole of Fe(NO3)3, the limiting reagent here is NaOH, which will produce 1.071 / 3 = 0.357 moles of Fe(OH)3. The mass of Fe(OH)3 precipitated is then 0.357 moles * 106.87 g/mol = 38.2 grams.

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