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.
I got 0.242 g Fe(OH)3 and 3.83 mmol Fe3+ remaining.
I agree with 3.83 mmols ferric but the problem asks for mols and not mmols.
I don't agree with 0.242 g Fe(OH)3 formed
for each of the following reactions 3.00 moles of each reactant is present initially. Determine the limiting reactant and calculate the moles of product in parentheses that would form
Fe2O3 +3H2=2Fe3H2O
To calculate the mass of Fe(OH)3(s) produced and the number of moles of the excess reactant remaining in solution, you need to follow a few steps. Here is how you can do it:
Step 1: Convert the given volumes of solutions to moles.
To convert the volumes of solutions to moles, you need to use the equation:
moles = concentration (M) × volume (L)
For KOH(aq):
moles of KOH = 0.153 M × (50/1000) L = 0.00765 mol
For Fe(NO3)3(aq):
moles of Fe(NO3)3 = 0.255 M × (25/1000) L = 0.006375 mol
Step 2: Determine the limiting reactant.
To find the limiting reactant, compare the stoichiometric ratios of the reactants in the balanced equation. The balanced equation for the reaction is:
3 KOH(aq) + Fe(NO3)3(aq) → Fe(OH)3(s) + 3 KNO3(aq)
From the balanced equation, you can see that the ratio of KOH to Fe(NO3)3 is 3:1. Therefore, the limiting reactant is Fe(NO3)3 since there is less of it compared to the stoichiometric ratio.
Step 3: Calculate the mass of Fe(OH)3(s) produced.
To calculate the mass of Fe(OH)3(s), you need to use the stoichiometric ratio between Fe(NO3)3 and Fe(OH)3. From the balanced equation, you can see that 1 mol of Fe(NO3)3 produces 1 mol of Fe(OH)3.
mass of Fe(OH)3 = moles of Fe(NO3)3 × molar mass of Fe(OH)3
The molar mass of Fe(OH)3 is:
(1 × molar mass of Fe) + (3 × molar mass of O) + (3 × molar mass of H)
Here are the molar masses of the elements:
molar mass of Fe = 55.845 g/mol
molar mass of O = 15.999 g/mol
molar mass of H = 1.008 g/mol
mass of Fe(OH)3 = 0.006375 mol × [(1 × 55.845) + (3 × 15.999) + (3 × 1.008)] g = 0.242 g
Therefore, the mass of Fe(OH)3(s) produced is 0.242 g.
Step 4: Calculate the number of moles of the excess reactant remaining in solution.
Since Fe(NO3)3 is the limiting reactant, we need to determine the number of moles of the excess reactant (KOH) remaining in solution.
moles of KOH remaining = moles of KOH initially - moles of KOH used
moles of KOH remaining = 0.00765 mol - (0.006375 mol × (3/1)) = 0.00765 mol - 0.019125 mol = -0.011475 mol
It seems that there is a negative value for moles of KOH remaining. This indicates that all of the KOH was consumed in the reaction, and there is no excess reactant left in the solution.
Therefore, the number of moles of the excess reactant remaining in solution is 0 mol.
In summary, the mass of Fe(OH)3(s) produced is 0.242 g, and there is no excess KOH remaining in solution. The remaining moles of Fe(NO3)3 are indeed 3.83 mmol.