Chemistry redox titration question
The question is "In a titration, 1.362 g of a moss killer was dissolved in 250 cm3 of distilled water. A 25 cm3 sample of the resulting solution reacted with 15.60 cm3 of acidified 0.0250 mol dm-3 KMnO4 solution. Calculate the percentage by mass of iron in this moss killer assuming that all of the iron is in the form of iron (II) ions"
The main concern i have is i have absoloutly no clue about the mole ratio.
The main concern i have is i have absoloutly no clue about the mole ratio.
Original post by Nitrotoluene
Hello smileybone001!😀
Note: mol dm-3 = mol/dm^3
Reactants:
Moss killer 1.362 g (contains iron(II) ions)
Acidified 0.0250 mol/dm^3 KMnO4 solution
Products:
Iron(II) ions
Manganese(II) ions (Mn^2-)
Water (H2O)
The redox reaction is:
From the problem, you know that 15.60 cm^3 of acidified 0.0250 mol/dm^3 KMnO4 solution was used. To find the number of moles of KMnO4, we can use the formula:
moles KMnO4 = 15.60 cm^3 x (1 dm^3 / 1000 cm^3) x 0.0250 mol/dm^3 = 0.0390 mol. Note: 1 dm^3 = 1000 cm^3 = 1 litre.
Now, you need to find the number of moles of iron(II) ions.
Since 25 cm^3 of the moss killer solution was used, we can assume that this volume contains a certain number of moles of iron(II) ions.
The molar mass of iron(II) ions is 55.85 g/mol. You can use this to find the number of moles of iron(II) ions in 1.362 g of the moss killer:
moles Fe^2+ = mass of moss killer (in g) / molar mass of iron(II) (in g/mol) = 1.362 g/ 55.85 g/mol = 0.0244 mol
Now, you can set up a mole ratio equation (mole ratio Fe^2+ : MnO4^ is 1 : 1 as per the balanced reaction above).
moles:
Substitute the values:
Solve for X:
So, the number of moles of iron(II) ions in 25 cm^3 of the moss killer solution is 0.0244 mol.
Now, you can calculate the percentage by mass of iron in the moss killer:
% mass = (moles Fe^2+ x molar mass Fe^2+)/ total mass = (0.0244 mol x 55.85 g/mol)/[(1.362 g + (250 cm^3 x 1 g/cm^3)] = 13.4%.
Note: 250 cm^3 = 250 g since the density of water is 1 g/cm^3; (250 cm^3 x 1 g/cm^3 = 250 g)
The percentage by mass of iron in the moss killer is approximately 13.4%.
Bye,
Sandro
Note: mol dm-3 = mol/dm^3
Reactants:
Moss killer 1.362 g (contains iron(II) ions)
Acidified 0.0250 mol/dm^3 KMnO4 solution
Products:
Iron(II) ions
Manganese(II) ions (Mn^2-)
Water (H2O)
The redox reaction is:
Fe^2+ + MnO4^- ==> Fe^3+ + Mn^2+ + H2O
From the problem, you know that 15.60 cm^3 of acidified 0.0250 mol/dm^3 KMnO4 solution was used. To find the number of moles of KMnO4, we can use the formula:
moles = volume (in dm^3) x concentration (in mol/dm^3)
moles KMnO4 = 15.60 cm^3 x (1 dm^3 / 1000 cm^3) x 0.0250 mol/dm^3 = 0.0390 mol. Note: 1 dm^3 = 1000 cm^3 = 1 litre.
Now, you need to find the number of moles of iron(II) ions.
Since 25 cm^3 of the moss killer solution was used, we can assume that this volume contains a certain number of moles of iron(II) ions.
You can call this number of moles "X".
The molar mass of iron(II) ions is 55.85 g/mol. You can use this to find the number of moles of iron(II) ions in 1.362 g of the moss killer:
moles Fe^2+ = mass of moss killer (in g) / molar mass of iron(II) (in g/mol) = 1.362 g/ 55.85 g/mol = 0.0244 mol
Now, you can set up a mole ratio equation (mole ratio Fe^2+ : MnO4^ is 1 : 1 as per the balanced reaction above).
moles:
Fe^2+ : moles KMnO4 = X : 0.0390 mol
Substitute the values:
0.0244 mol : 0.0390 mol = X : 0.0390 mol
Solve for X:
X = (0.0244 x 0.0390)/0.0390
After simplifying the equation, you will have:
X = 0.0244
After simplifying the equation, you will have:
X = 0.0244
So, the number of moles of iron(II) ions in 25 cm^3 of the moss killer solution is 0.0244 mol.
Now, you can calculate the percentage by mass of iron in the moss killer:
% mass = (moles Fe^2+ x molar mass Fe^2+)/ total mass = (0.0244 mol x 55.85 g/mol)/[(1.362 g + (250 cm^3 x 1 g/cm^3)] = 13.4%.
Note: 250 cm^3 = 250 g since the density of water is 1 g/cm^3; (250 cm^3 x 1 g/cm^3 = 250 g)
The percentage by mass of iron in the moss killer is approximately 13.4%.
Errors and omissions excepted!
Bye,
Sandro
Thank you for the response!
I understand a lot better now but i'm still not able to get the 13.4% outcome you achieved, I think im getting hung up on the % mass.
Original post by smileybone001
Thank you for the response!
I understand a lot better now but i'm still not able to get the 13.4% outcome you achieved, I think im getting hung up on the % mass.
I understand a lot better now but i'm still not able to get the 13.4% outcome you achieved, I think im getting hung up on the % mass.
Disregard my answer because I have not balanced the redox reaction.
This is the well-balanced reaction: 5 Fe^2+ + MnO4^- + 8 H^+ = 5 Fe^3+ + Mn^2+ + 4 H2O.
I apologise for my serious inattention.
Sandro
Original post by Nitrotoluene
Hello smileybone001!😀
Note: mol dm-3 = mol/dm^3
Reactants:
Moss killer 1.362 g (contains iron(II) ions)
Acidified 0.0250 mol/dm^3 KMnO4 solution
Products:
Iron(II) ions
Manganese(II) ions (Mn^2-)
Water (H2O)
The redox reaction is:
Now, you need to find the number of moles of iron(II) ions.
Since 25 cm^3 of the moss killer solution was used, we can assume that this volume contains a certain number of moles of iron(II) ions.
moles Fe^2+ = mass of moss killer (in g) / molar mass of iron(II) (in g/mol) = 1.362 g/ 55.85 g/mol = 0.0244 mol
Now, you can set up a mole ratio equation (mole ratio Fe^2+ : MnO4^ is 1 : 1 as per the balanced reaction above).
moles:
Now, you can calculate the percentage by mass of iron in the moss killer:
% mass = (moles Fe^2+ x molar mass Fe^2+)/ total mass = (0.0244 mol x 55.85 g/mol)/[(1.362 g + (250 cm^3 x 1 g/cm^3)] = 13.4%.
Note: 250 cm^3 = 250 g since the density of water is 1 g/cm^3; (250 cm^3 x 1 g/cm^3 = 250 g)
The percentage by mass of iron in the moss killer is approximately 13.4%.
Sandro
Note: mol dm-3 = mol/dm^3
Reactants:
Moss killer 1.362 g (contains iron(II) ions)
Acidified 0.0250 mol/dm^3 KMnO4 solution
Products:
Iron(II) ions
Manganese(II) ions (Mn^2-)
Water (H2O)
The redox reaction is:
Fe^2+ + MnO4^- ==> Fe^3+ + Mn^2+ + H2O
From the problem, you know that 15.60 cm^3 of acidified 0.0250 mol/dm^3 KMnO4 solution was used. To find the number of moles of KMnO4, we can use the formula:moles = volume (in dm^3) x concentration (in mol/dm^3)
moles KMnO4 = 15.60 cm^3 x (1 dm^3 / 1000 cm^3) x 0.0250 mol/dm^3 = 0.0390 mol. Note: 1 dm^3 = 1000 cm^3 = 1 litre.Now, you need to find the number of moles of iron(II) ions.
Since 25 cm^3 of the moss killer solution was used, we can assume that this volume contains a certain number of moles of iron(II) ions.
You can call this number of moles "X".
The molar mass of iron(II) ions is 55.85 g/mol. You can use this to find the number of moles of iron(II) ions in 1.362 g of the moss killer:moles Fe^2+ = mass of moss killer (in g) / molar mass of iron(II) (in g/mol) = 1.362 g/ 55.85 g/mol = 0.0244 mol
Now, you can set up a mole ratio equation (mole ratio Fe^2+ : MnO4^ is 1 : 1 as per the balanced reaction above).
moles:
Fe^2+ : moles KMnO4 = X : 0.0390 mol
Substitute the values:0.0244 mol : 0.0390 mol = X : 0.0390 mol
Solve for X:X = (0.0244 x 0.0390)/0.0390
After simplifying the equation, you will have:
X = 0.0244
So, the number of moles of iron(II) ions in 25 cm^3 of the moss killer solution is 0.0244 mol.After simplifying the equation, you will have:
X = 0.0244
Now, you can calculate the percentage by mass of iron in the moss killer:
% mass = (moles Fe^2+ x molar mass Fe^2+)/ total mass = (0.0244 mol x 55.85 g/mol)/[(1.362 g + (250 cm^3 x 1 g/cm^3)] = 13.4%.
Note: 250 cm^3 = 250 g since the density of water is 1 g/cm^3; (250 cm^3 x 1 g/cm^3 = 250 g)
The percentage by mass of iron in the moss killer is approximately 13.4%.
Errors and omissions excepted!
Bye,Sandro
Dear smileybone001!
I humbly apologise for the mess I made.
I will now remedy this by giving you the carefully re-read solution to the problem.
Here it is!
To calculate the percentage by mass of iron in the moss killer, you shall follow these steps:
You have to calculate the moles of KMnO4 used in the reaction
1. Concentration of KMnO4 solution = 0.0250 mol/dm^3
2. Volume of KMnO4 used = 15.60 cm^3 = 0.01560 dm^3
Using the formula for moles when you know concentration and volume:
Moles of KMnO4 = Concentration KMnO4 x Volume KMnO4
Moles of KMnO4 = 0.0250 mol/dm^3 x 0.01560 dm^3 = 0.000390 mol
Use the stoichiometry of the reaction to find moles of Fe^2+
From the balanced equation:
5 Fe^2+ + MnO4^- + 8 H^+ = 5 Fe^3+ + Mn^2+ + 4 H2O
You can see that 1 mole of KMnO4 reacts with 5 moles of Fe^2+.
The molar ratio of reaction Fe^2+ : MnO4^- is 5 : 1.
You can see that 1 mole of KMnO4 reacts with 5 moles of Fe^2+.
The molar ratio of reaction Fe^2+ : MnO4^- is 5 : 1.
Hence, the moles of Fe^2+ can be calculated as follows:
Moles of Fe^2+ = (5 x Moles of MnO4^-) = (5 x 0.000390 mol) = 0.001950 mol
Now you have to calculate the mass of Fe^2+ in the sample.
The molar mass of iron (Fe^2+) is 55.85 g/mol.
Now, you can find the mass of Fe^2+:
Mass of Fe^2+ = Moles Fe^2+ x Molar Mass Fe^2+
Mass of Fe^2+ = 0.001950 mol x 55.85 g/mol = 0.109 g
Now, you can calculate the total mass of iron in the entire solution
Since the 25 cm^3 sample is part of the total 250 cm^3 solution (or better you took 25 cm^3 from 250 cm^3 of the moss killer solution), you can find the total mass of iron in the entire solution:
Total mass of Fe^2+ = (0.109 g/25 cm^3) x 250 cm^3 = 1.09 g
Calculate the percentage by mass of iron in the moss killer.
Now you can find the percentage by mass of iron in the moss killer:
Percentage by mass of Fe = (Total mass of Fe^2+/Mass of moss killer) x 100
Where:
-Mass of moss killer = 1.362 g
Percentage by mass of Fe = (1.09 g/1.362 g) x 100 = 80.02%
Conclusion
Conclusion
The percentage by mass of iron in the moss killer is approximately 80.02%.
Bye,
Sandro
(edited 3 months ago)
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