Calculate the mass of Fe(OH)3 s) produced by mixing 50.0 mL of 0.153 M KOH (aq) and 25.0 mL of 0.255 M Fe(NO3)3 (aq), and the number of moles of the excess reactant remaining in solution.
To calculate the mass of Fe(OH)3(s) produced, we first need to determine the limiting reactant. This is the reactant that will be completely consumed, limiting the amount of product that can be formed.
Let's start by calculating the number of moles of each reactant:
For KOH:
Number of moles of KOH = volume (in L) × concentration
Number of moles of KOH = (50.0 mL ÷ 1000 mL/L) × 0.153 M
Number of moles of KOH = 0.00765 moles
For Fe(NO3)3:
Number of moles of Fe(NO3)3 = volume (in L) × concentration
Number of moles of Fe(NO3)3 = (25.0 mL ÷ 1000 mL/L) × 0.255 M
Number of moles of Fe(NO3)3 = 0.006375 moles
Now, we need to determine the stoichiometry of the reaction from the balanced chemical equation to find the ratio of the reactants and products.
The balanced chemical equation for the reaction between KOH and Fe(NO3)3 is:
3 KOH + Fe(NO3)3 → Fe(OH)3 + 3 KNO3
According to the balanced equation, the mole ratio of KOH to Fe(OH)3 is 3:1.
Since the mole ratio of KOH to Fe(OH)3 is 3:1, the number of moles of Fe(OH)3 produced will be 0.00765 moles × (1 mole Fe(OH)3 / 3 moles KOH) = 0.00255 moles
To calculate the mass of Fe(OH)3 produced, multiply the number of moles by the molar mass of Fe(OH)3.
The molar mass of Fe(OH)3 is:
Fe: 55.845 g/mol
O: 16.00 g/mol × 3 = 48.00 g/mol
H: 1.008 g/mol × 3 = 3.024 g/mol
Total molar mass of Fe(OH)3 = 55.845 g/mol + 48.00 g/mol + 3.024 g/mol = 106.869 g/mol
Mass of Fe(OH)3 produced = number of moles × molar mass
Mass of Fe(OH)3 produced = 0.00255 moles × 106.869 g/mol ≈ 0.273 g
Now, let's determine the excess reactant. The reactant that is not completely consumed is the excess reactant.
To find the moles of the excess reactant remaining in solution, we compare the stoichiometry of the balanced equation with the number of moles used for each reactant.
For KOH:
Number of moles of KOH used = 0.00765 moles
Number of moles of KOH needed (according to stoichiometry) = 0.00765 moles × (3 moles KOH/1 mole Fe(OH)3) = 0.02295 moles
Number of moles of KOH remaining = 0.02295 moles - 0.00765 moles = 0.0153 moles
For Fe(NO3)3:
Number of moles of Fe(NO3)3 used = 0.006375 moles
Number of moles of Fe(NO3)3 needed (according to stoichiometry) = 0.006375 moles × (1 mole Fe(NO3)3/1 mole Fe(OH)3) = 0.006375 moles
Number of moles of Fe(NO3)3 remaining = 0.006375 moles - 0.006375 moles = 0 moles
Therefore, the excess reactant is KOH, and the number of moles of KOH remaining in solution is 0.0153 moles.
In summary:
- Mass of Fe(OH)3(s) produced = 0.273 g
- Number of moles of KOH remaining = 0.0153 moles
To calculate the mass of Fe(OH)3 produced, we first need to determine the limiting reactant. The limiting reactant is the one that is fully consumed and determines the amount of product that can be formed.
1. Convert the volumes to moles:
- For KOH:
Number of moles of KOH = volume (L) × molarity (mol/L)
Number of moles of KOH = 0.0500 L × 0.153 mol/L
- For Fe(NO3)3:
Number of moles of Fe(NO3)3 = volume (L) × molarity (mol/L)
Number of moles of Fe(NO3)3 = 0.0250 L × 0.255 mol/L
2. Determine the mole ratio between KOH and Fe(OH)3:
The balanced chemical equation for the reaction between KOH and Fe(NO3)3 is:
3 KOH + Fe(NO3)3 → Fe(OH)3 + 3 KNO3
According to the balanced equation, the mole ratio between KOH and Fe(OH)3 is 3:1.
3. Calculate the moles of Fe(OH)3 that can be formed from each reactant:
Divide the number of moles of each reactant by their respective coefficients in the balanced equation.
Moles of Fe(OH)3 from KOH = Moles of KOH × (1 mole Fe(OH)3 / 3 moles KOH)
Moles of Fe(OH)3 from Fe(NO3)3 = Moles of Fe(NO3)3 × (1 mole Fe(OH)3 / 1 mole Fe(NO3)3)
4. Compare the moles of Fe(OH)3 obtained from each reactant:
The reactant that produces fewer moles of Fe(OH)3 is the limiting reactant.
5. Calculate the mass of Fe(OH)3 produced:
Once you identify the limiting reactant, multiply its moles by the molar mass of Fe(OH)3 to obtain the mass.
To determine the number of moles of the excess reactant remaining in solution, use the moles of the excess reactant and subtract from the original moles of the reactant initially present in solution.
Note: Make sure to check the calculations and use appropriate significant figures throughout the calculations.
This is a limiting reagent (LR) problem. You know that because an amount is given for both reactants.
3KOH + Fe(NO3)3 ==> Fe(OH)3 + 3KNO3
mols KOH = M x L = ?
mols Fe(NO3)3 = M x L = ?
Using the coefficients in the balanced equation, convert mols KOH to mols Fe(OH)3.
Do the same to convert mols Fe(NO3)3 to mols Fe(OH)3.
It is likely that these two values for Fe(OH)3 will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that smaller number is the LR. (The other (excess) reagent I will call the ER.)
To fine grams Fe(OH)3 it is grams = mols Fe(OH)3 x molar mass Fe(OH)3.
Now to find mols of ER. That is convert mols of LR used (all of it) to mols ER used. Then mols ER remaining = initial mols ER - mols ER used = ?