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 1 month ago)

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