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.

Sure! I'd be happy to help you with these questions. Let's break it down step by step:

1) To find the mass of Fe(OH)3 that should precipitate, we first need to determine the limiting reactant. The stoichiometry of the balanced chemical equation will help us with this:

FeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl

From the equation, we can see that the ratio of FeCl3 to Fe(OH)3 is 1:1. That means that for every 1 mole of FeCl3, 1 mole of Fe(OH)3 is produced.

First, let's find the moles of FeCl3 and NaOH:

Moles of FeCl3 = Volume (L) × Molarity (mol/L)
= 20.2 mL × (1 L / 1000 mL) × 0.234 mol/L
= 0.0047376 mol

Moles of NaOH = Volume (L) × Molarity (mol/L)
= 42.5 mL × (1 L / 1000 mL) × 0.453 mol/L
= 0.01925625 mol

Since the ratio of FeCl3 to Fe(OH)3 is 1:1, the moles of Fe(OH)3 that should precipitate will also be 0.0047376 mol.

To find the mass of Fe(OH)3, we can use its molar mass:

Mass of Fe(OH)3 = Moles × Molar Mass
= 0.0047376 mol × 106.867 g/mol
≈ 0.505 g

Therefore, approximately 0.505 grams of Fe(OH)3 should precipitate (assuming 100% yield).

2) To find the moles of the excess reactant remaining after the reaction is complete, we need to determine the limiting reactant first (which we found to be FeCl3). Since FeCl3 is completely consumed, NaOH is the excess reactant.

The balanced chemical equation shows that 3 moles of NaOH react with 1 mole of FeCl3 to produce 1 mole of Fe(OH)3. Therefore, the number of moles of NaOH that reacted can be calculated using the mole ratio:

Moles of NaOH reacted = Moles of FeCl3 × (3 moles NaOH / 1 mole FeCl3)
= 0.0047376 mol × (3 / 1)
= 0.0142128 mol

To find the moles of the excess reactant remaining, we subtract the moles reacted from the initial moles of NaOH:

Moles of excess NaOH remaining = Initial moles of NaOH - Moles of NaOH reacted
= 0.01925625 mol - 0.0142128 mol
= 0.00504345 mol

Therefore, approximately 0.00504345 moles of NaOH will remain after the reaction is complete.

3) Finally, to find the molar concentration of the excess reactant at the end of the reaction, we need to know the final volume of the solution. For this, we add the initial volumes of FeCl3 and NaOH:

Total volume of solution = Volume of FeCl3 + Volume of NaOH
= 20.2 mL + 42.5 mL
= 62.7 mL
= 0.0627 L

The molar concentration (M) is given by:

Molarity = Moles of solute / Volume of solution (in liters)

Molarity of the excess NaOH = Moles of excess NaOH remaining / Total volume of solution
= 0.00504345 mol / 0.0627 L
≈ 0.0804 M

Therefore, the molar concentration of the excess NaOH at the end of the reaction is approximately 0.0804 M.

I hope this explanation helps! Let me know if you have any further questions.