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:
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.
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?
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