Problem: Consider this reaction:Fe(NO3)3(aq) + NaOH(aq) → Fe(OH)3(s) + 3 NaNO3(aq)If there is 25.00 mL of 0.650 M Fe(NO3)3 and 20.00 mL of 1.014 M NaOH, how many grams of iron(III) hydroxide theoretically form?

We’re being asked to calculate the mass of iron(III) hydroxide theoretically form (Fe(OH)₃) in the reaction of 25.00 mL of 0.650 M Fe(NO₃)₃ and 20.00 mL of 1.014 M NaOH.

Recall that molarity is the ratio of the moles of solute and the volume of solution (in liters). 

In other words:

$\boxed{\mathbf{Molarity}\mathbf{ }\mathbf{\left(}\mathbf{M}\mathbf{\right)}\mathbf{=}\frac{\mathbf{moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{solute}}{\mathbf{Liters}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{solution}}}$

Notice that we are given the volume and molarity of both reactants: 

this means we need to determine the limiting reactant, which is the reactant that forms the less amount of product.