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

Explanation:

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In this case, given the chemical reaction:

In such a way, given the volumes and molarities of each reactant, we can compute the moles of produced iron (III) hydroxide by each of them, via the 3:1 and 1:1 mole ratios:

It means that the sodium hydroxide is the limiting reactant and 0.00833 moles of iron (III) hydroxide are produced; thus, the required mass is:

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