How many grams of iron(III) hydroxide (106.87 g/mol) will precipitate if 50.0 mL of 0.500 M sodium hydroxide is added to 75.0 mL of 0.200 M iron(III) nitrate?

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Quizplus · Accepted Answer

The reaction is Fe(NO3)3 + 3NaOH → Fe(OH)3 + 3NaNO3. Moles of Fe(NO3)3 = 0.075 L * 0.200 M = 0.015 mol. Moles of NaOH = 0.050 L * 0.500 M = 0.025 mol. NaOH is the limiting reagent. Moles of Fe(OH)3 = 0.025 mol NaOH * (1 mol Fe(OH)3 / 3 mol NaOH) = 0.00833 mol. Mass of Fe(OH)3 = 0.00833 mol * 106.87 g/mol = 0.891 g.

How Many Grams of Iron(III) Hydroxide (106