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?

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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:


Molarity (M)=moles of soluteLiters of 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


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Problem Details

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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