posted by Lilly on .
20.2 mL of 0.234 M FeCl3 is added to 42.5 mL of 0.453 M NaOH. Iron(III) hydroxide precipitates. NaCl remains in aqueous solution.
1) What mass (in g) of Fe(OH)3 should precipitate (assuming 100% yield)?
2) How many moles of the excess reactant remain after the reaction is complete?
3) What is the molor concentration of the excess reactant at the end of the reaction? (M = moles solute / liters of solution)
I'd appreciate help.
This is a limiting reagent problem. How do I know that (in addition to the problem saying so)? Because BOTH reactants are given.
1. Write the equation and balance it.
2. Convert FeCl3 to moles. moles = M x L.
3. Convert NaOH to moles. same procedure.
4. Using the coefficients in the balanced equation, convert moles FeCl3 to moles Fe(OH)3.
5. Same procedure, convert moles NaOH to moles Fe(OH)3.
6. It is most likely that the answers from steps 4 and 5 will not agree which means one of them is wrong. The correct value, in limiting reagent problems, is ALWAYS the smaller value and the reagent producing that number is the limiting reagent.
7. Now convert the smaller value to grams. g = moles x molar mass. That is the answer for your question 2.
#2. moles of excess reagent remaining unreacted. Using the coefficients in the balanced equation, convert moles of the limiting reagent to moles of the non-limiting reagent.
3. M = excss moles of non-limiting reagent(answer from 2)/total volume.