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 0.00383 mol Fe(NO3)3 remaining.
I don't know how to obtain the mass of Fe(OH)3, but here is my work:
(0.00255 mol KOH) x (1mol Fe(OH)3/3mol KOH) x (106.88 g/1mol Fe(OH)3
The answer key says 0.273 g Fe(OH)3 produced ( when I calculate the work is set up, this answer appeared when I did not divide it by 3, but there are 3 moles of KOH for every 1 mol of Fe(OH)3.)??
your error, I think, is you divided twice.
50ml of .153M is .00765mol KOH. That is three times as much as in your KOH used.
.00765*1/3*106.88=.273g
Thank you!
To calculate the mass of Fe(OH)3 produced, you need to use the mole-to-mole ratio between KOH and Fe(OH)3, as well as the molar mass of Fe(OH)3.
First, let's calculate the moles of KOH used:
Moles of KOH = Volume of KOH solution (L) x Molarity of KOH (mol/L)
= 0.0500 L x 0.153 mol/L
= 0.00765 mol KOH
Next, use the mole-to-mole ratio between KOH and Fe(OH)3:
Moles of Fe(OH)3 = Moles of KOH x (1 mol Fe(OH)3 / 3 mol KOH)
= 0.00765 mol KOH x (1 mol Fe(OH)3 / 3 mol KOH)
= 0.00255 mol Fe(OH)3
Finally, calculate the mass of Fe(OH)3:
Mass of Fe(OH)3 = Moles of Fe(OH)3 x Molar mass of Fe(OH)3
= 0.00255 mol Fe(OH)3 x 106.88 g/mol Fe(OH)3
= 0.273 g Fe(OH)3
So, the mass of Fe(OH)3 produced is 0.273 g.
Regarding the remaining excess reactant, Fe(NO3)3, your calculation of 0.00383 mol Fe(NO3)3 remaining seems correct.
To calculate the mass of Fe(OH)3 produced, you need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
First, let's find the moles of Fe(NO3)3 and KOH present:
Moles of Fe(NO3)3 = volume (L) x molarity (mol/L) = 0.025 L x 0.255 mol/L = 0.006375 mol Fe(NO3)3
Moles of KOH = volume (L) x molarity (mol/L) = 0.05 L x 0.153 mol/L = 0.00765 mol KOH
To determine which is the limiting reagent, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation:
Fe(NO3)3 + 3KOH -> Fe(OH)3 + 3KNO3
From the balanced equation, we can see that 1 mole of Fe(NO3)3 reacts with 3 moles of KOH to produce 1 mole of Fe(OH)3.
Moles of Fe(NO3)3 remaining = Moles of Fe(NO3)3 - (Moles of KOH / Reaction stoichiometry)
= 0.006375 mol - (0.00765 mol / 3)
= 0.006375 mol - 0.00255 mol
= 0.003825 mol Fe(NO3)3 remaining
So, your calculation for the remaining moles of Fe(NO3)3 is correct: 0.003825 mol.
Next, to calculate the mass of Fe(OH)3 produced, use the moles of KOH remaining:
Moles of KOH remaining = Moles of KOH - (Moles of Fe(NO3)3 remaining x Reaction stoichiometry)
= 0.00765 mol - (0.003825 mol x 3)
= 0.00765 mol - 0.011475 mol
= -0.003825 mol
Since we cannot have a negative amount of a substance, this means that KOH is completely consumed and Fe(NO3)3 is in excess. Therefore, the moles of Fe(OH)3 produced is equal to the moles of KOH used:
Moles of Fe(OH)3 produced = Moles of KOH
= 0.00765 mol
Now, to obtain the mass of Fe(OH)3 produced, use the molar mass of Fe(OH)3:
Mass of Fe(OH)3 = Moles of Fe(OH)3 x Molar mass of Fe(OH)3
= 0.00765 mol x 106.88 g/mol
= 0.817 g
So, the correct mass of Fe(OH)3 produced is approximately 0.817 g, not 0.273 g as mentioned in the answer key.
To summarize:
- The remaining moles of Fe(NO3)3 is calculated correctly as 0.003825 mol.
- The correct mass of Fe(OH)3 produced is approximately 0.817 g.