Hey guys

ive been set some questions to do on redox titration calculations , but the two im stuck on are ....

3) 3.00 g of a lawn sand containing an iron (II) salt was shaken with dilute H2SO4. The resulting solution required 25.00 cm3 of 0.0200 mol dm-3 potassium manganate (VII) to oxidise the Fe2+ ions in the solution to Fe3+ ions. Use this to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand.

4) Ammonium iron (II) sulphate crystals have the following formula: (NH4)2SO4.FeSO4.nH2O. In an experiment to find n, 8.492 g of the salt were dissolved and made up to 250 cm3 solution with distilled water and dilute sulphuric acid. A 25.0 cm3 portion of the solution was titrated against 0.0150 mol dm-3 KMnO4, 22.5 cm3 being required. Calculate n.

i would find it incredibly useful if someone could talk me through what i have to do in order to solve these questions (in detail please!)

i need to know how to do these ones to do the rest of the questions ive been set aswell, so you can use these ones as examples to show me how they can be done

thanks !

ive been set some questions to do on redox titration calculations , but the two im stuck on are ....

3) 3.00 g of a lawn sand containing an iron (II) salt was shaken with dilute H2SO4. The resulting solution required 25.00 cm3 of 0.0200 mol dm-3 potassium manganate (VII) to oxidise the Fe2+ ions in the solution to Fe3+ ions. Use this to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand.

4) Ammonium iron (II) sulphate crystals have the following formula: (NH4)2SO4.FeSO4.nH2O. In an experiment to find n, 8.492 g of the salt were dissolved and made up to 250 cm3 solution with distilled water and dilute sulphuric acid. A 25.0 cm3 portion of the solution was titrated against 0.0150 mol dm-3 KMnO4, 22.5 cm3 being required. Calculate n.

i would find it incredibly useful if someone could talk me through what i have to do in order to solve these questions (in detail please!)

i need to know how to do these ones to do the rest of the questions ive been set aswell, so you can use these ones as examples to show me how they can be done

thanks !

LalTheBlondeOne

Hey guys

ive been set some questions to do on redox titration calculations , but the two im stuck on are ....

3) 3.00 g of a lawn sand containing an iron (II) salt was shaken with dilute H2SO4. The resulting solution required 25.00 cm3 of 0.0200 mol dm-3 potassium manganate (VII) to oxidise the Fe2+ ions in the solution to Fe3+ ions. Use this to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand.

ive been set some questions to do on redox titration calculations , but the two im stuck on are ....

3) 3.00 g of a lawn sand containing an iron (II) salt was shaken with dilute H2SO4. The resulting solution required 25.00 cm3 of 0.0200 mol dm-3 potassium manganate (VII) to oxidise the Fe2+ ions in the solution to Fe3+ ions. Use this to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand.

1. Work out moles of manganate (VII) ions

2. Use the equation for the reaction between iron (II) and manganate (VII) ions to get the moles equivalent of iron (II)

3. Multiply moles of iron (II) by RAM of iron to get mass of iron

4. mass iron/total mass x 100 = percentage by mass of iron in lawnsand

LalTheBlondeOne

4) Ammonium iron (II) sulphate crystals have the following formula: (NH4)2SO4.FeSO4.nH2O. In an experiment to find n, 8.492 g of the salt were dissolved and made up to 250 cm3 solution with distilled water and dilute sulphuric acid. A 25.0 cm3 portion of the solution was titrated against 0.0150 mol dm-3 KMnO4, 22.5 cm3 being required. Calculate n.

i would find it incredibly useful if someone could talk me through what i have to do in order to solve these questions (in detail please!)

i need to know how to do these ones to do the rest of the questions ive been set aswell, so you can use these ones as examples to show me how they can be done

thanks !

4) Ammonium iron (II) sulphate crystals have the following formula: (NH4)2SO4.FeSO4.nH2O. In an experiment to find n, 8.492 g of the salt were dissolved and made up to 250 cm3 solution with distilled water and dilute sulphuric acid. A 25.0 cm3 portion of the solution was titrated against 0.0150 mol dm-3 KMnO4, 22.5 cm3 being required. Calculate n.

i would find it incredibly useful if someone could talk me through what i have to do in order to solve these questions (in detail please!)

i need to know how to do these ones to do the rest of the questions ive been set aswell, so you can use these ones as examples to show me how they can be done

thanks !

The second question is very similar

LalTheBlondeOne

3) 3.00 g of a lawn sand containing an iron (II) salt was shaken with dilute H2SO4. The resulting solution required 25.00 cm3 of 0.0200 mol dm-3 potassium manganate (VII) to oxidise the Fe2+ ions in the solution to Fe3+ ions. Use this to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand.

3) 3.00 g of a lawn sand containing an iron (II) salt was shaken with dilute H2SO4. The resulting solution required 25.00 cm3 of 0.0200 mol dm-3 potassium manganate (VII) to oxidise the Fe2+ ions in the solution to Fe3+ ions. Use this to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand.

1.) First write the redox equation:

MnO4- + 8H+ + 5Fe2+ -------> Mn2+ + 5Fe3+ + 4H2O

2.) Work out the moles of what you know (in this case the manganate)

so n= conc * V(cm^3)/1000

n = 0.0200 * 25/1000 = 5.00*10^-4 mol of MnO4-

3.) Next use ratios to find moles of Fe2+

MnO4- : Fe2+

1:5

therefore moles of Fe2+ = moles of MnO4- * 5

.... which = 2.50*10^-3

4.) Work out grams of Fe2+ in the sample

grams = n * Mr = (2.50*10^-3) * 55.8

= 0.140g

5.) Finally work out % by mass

so it's = (grams of Fe2+ in sample) / (grams of lawn sand) all * 100

= (0.140/3.00) * 100

= 4.67% (2dp)

LalTheBlondeOne

4) Ammonium iron (II) sulphate crystals have the following formula: (NH4)2SO4.FeSO4.nH2O. In an experiment to find n, 8.492 g of the salt were dissolved and made up to 250 cm3 solution with distilled water and dilute sulphuric acid. A 25.0 cm3 portion of the solution was titrated against 0.0150 mol dm-3 KMnO4, 22.5 cm3 being required. Calculate n.

1) again first write the redox equation:

MnO4- + 8H+ + 5Fe2+ -------> Mn2+ + 5Fe3+ + 4H2O

2) Work out Moles of what you know (Manganate again)

n= conc * V(cm^3) / 1000

= 0.0150* 22.5/1000

=3.38*10^-4 mol of MnO4-

3.) next use ratios to find moles of Fe2+ in 25cm^3

MnO4- : Fe2+

1:5

therefore moles of Fe2+ = moles of MnO4- * 5

= (3.38*10^-4) * 5

= 1.69*10^-3 mol of Fe2+ in 25cm^3

4.) work out moles of Fe2+ in the original 250cm^3 solution

..... 25 * 10 = 250

therefore (1.69*10^-3) * 10 = 0.0169mol of Fe2+ in the 250cm^3 solution

5.) Moles of Fe2+ are equal to the moles of the whole salt , as for every unit of (NH4)2SO4.FeSO4.nH2O there is one unit of FeSO4

6.) we can therefore work out the Mr of the salt

Mr = grams/moles

= 8.492 / (0/0169)

= 502(3sf)

7.) Next work ot the Mr of (NH4)2SO4.FeSO4 using your periodic table

so for N (2*14)

for H (8*1)

for S (2*32.1)

for O (8 * 16)

for Fe(1*55.8)

TOTAL = 284

So minus 284 form the total Mr of the salt to gwt the Mr of the water of crystallisation

502-284 = 218

Mr of 1 molecule of H2O is (1*2)+(1*16)= 18

therefore number of water molecule (n)

...... = 218/18

= 12.11 (2dp)

So value of n is 12!!

Phew! hope that explained it xxx

Chem1993

Phew! hope that explained it xxx

Phew! hope that explained it xxx

You didn't actually leave the OP much thinking to do for himself...

charco

You didn't actually leave the OP much thinking to do for himself...

Maybe, in hindsight, I shouldn't have given him quite so much help. But i know alot of people find it helpful when their shown every bit step by step - particulary if their struggling. But I do understand where you coming from

herself but thank you x

mole of K2Cr2O7: 16.6/1000 * 0.22=3.652*10^-4

mole ratio: K2Cr2O7:Fe2+=1:5

mole of Fe2+ in 25cm-3: 1.826*10^-3

mole of Fe2+ in 1dm-3: (1.826*10^-3) * 1000/25= 0.07304

mass of Fe: ( 55.8+32.1+16*4)*0.07304=11.095

mass of H2O= 24.4-11.095=13.3

mass ratio between H2O and the whole crystal=13.3:24.4

n=

18n/(18n+55.8+32.1+16.4)=13.3/24.4

n=10

mole ratio: K2Cr2O7:Fe2+=1:5

mole of Fe2+ in 25cm-3: 1.826*10^-3

mole of Fe2+ in 1dm-3: (1.826*10^-3) * 1000/25= 0.07304

mass of Fe: ( 55.8+32.1+16*4)*0.07304=11.095

mass of H2O= 24.4-11.095=13.3

mass ratio between H2O and the whole crystal=13.3:24.4

n=

18n/(18n+55.8+32.1+16.4)=13.3/24.4

n=10

Original post by Chem1993

Phew! hope that explained it xxx

Phew! hope that explained it xxx

i was stuck on step 5 and knew it was something like this, but you setting it out really helped, thanks so much! x

Original post by Chem1993

1.) First write the redox equation:

MnO4- + 8H+ + 5Fe2+ -------> Mn2+ + 5Fe3+ + 4H2O

2.) Work out the moles of what you know (in this case the manganate)

so n= conc * V(cm^3)/1000

n = 0.0200 * 25/1000 = 5.00*10^-4 mol of MnO4-

3.) Next use ratios to find moles of Fe2+

MnO4- : Fe2+

1:5

therefore moles of Fe2+ = moles of MnO4- * 5

.... which = 2.50*10^-3

4.) Work out grams of Fe2+ in the sample

grams = n * Mr = (2.50*10^-3) * 55.8

= 0.140g

5.) Finally work out % by mass

so it's = (grams of Fe2+ in sample) / (grams of lawn sand) all * 100

= (0.140/3.00) * 100

= 4.67% (2dp)

1) again first write the redox equation:

MnO4- + 8H+ + 5Fe2+ -------> Mn2+ + 5Fe3+ + 4H2O

2) Work out Moles of what you know (Manganate again)

n= conc * V(cm^3) / 1000

= 0.0150* 22.5/1000

=3.38*10^-4 mol of MnO4-

3.) next use ratios to find moles of Fe2+ in 25cm^3

MnO4- : Fe2+

1:5

therefore moles of Fe2+ = moles of MnO4- * 5

= (3.38*10^-4) * 5

= 1.69*10^-3 mol of Fe2+ in 25cm^3

4.) work out moles of Fe2+ in the original 250cm^3 solution

..... 25 * 10 = 250

therefore (1.69*10^-3) * 10 = 0.0169mol of Fe2+ in the 250cm^3 solution

5.) Moles of Fe2+ are equal to the moles of the whole salt , as for every unit of (NH4)2SO4.FeSO4.nH2O there is one unit of FeSO4

6.) we can therefore work out the Mr of the salt

Mr = grams/moles

= 8.492 / (0/0169)

= 502(3sf)

7.) Next work ot the Mr of (NH4)2SO4.FeSO4 using your periodic table

so for N (2*14)

for H (8*1)

for S (2*32.1)

for O (8 * 16)

for Fe(1*55.8)

TOTAL = 284

So minus 284 form the total Mr of the salt to gwt the Mr of the water of crystallisation

502-284 = 218

Mr of 1 molecule of H2O is (1*2)+(1*16)= 18

therefore number of water molecule (n)

...... = 218/18

= 12.11 (2dp)

So value of n is 12!!

Phew! hope that explained it xxx

MnO4- + 8H+ + 5Fe2+ -------> Mn2+ + 5Fe3+ + 4H2O

2.) Work out the moles of what you know (in this case the manganate)

so n= conc * V(cm^3)/1000

n = 0.0200 * 25/1000 = 5.00*10^-4 mol of MnO4-

3.) Next use ratios to find moles of Fe2+

MnO4- : Fe2+

1:5

therefore moles of Fe2+ = moles of MnO4- * 5

.... which = 2.50*10^-3

4.) Work out grams of Fe2+ in the sample

grams = n * Mr = (2.50*10^-3) * 55.8

= 0.140g

5.) Finally work out % by mass

so it's = (grams of Fe2+ in sample) / (grams of lawn sand) all * 100

= (0.140/3.00) * 100

= 4.67% (2dp)

1) again first write the redox equation:

MnO4- + 8H+ + 5Fe2+ -------> Mn2+ + 5Fe3+ + 4H2O

2) Work out Moles of what you know (Manganate again)

n= conc * V(cm^3) / 1000

= 0.0150* 22.5/1000

=3.38*10^-4 mol of MnO4-

3.) next use ratios to find moles of Fe2+ in 25cm^3

MnO4- : Fe2+

1:5

therefore moles of Fe2+ = moles of MnO4- * 5

= (3.38*10^-4) * 5

= 1.69*10^-3 mol of Fe2+ in 25cm^3

4.) work out moles of Fe2+ in the original 250cm^3 solution

..... 25 * 10 = 250

therefore (1.69*10^-3) * 10 = 0.0169mol of Fe2+ in the 250cm^3 solution

5.) Moles of Fe2+ are equal to the moles of the whole salt , as for every unit of (NH4)2SO4.FeSO4.nH2O there is one unit of FeSO4

6.) we can therefore work out the Mr of the salt

Mr = grams/moles

= 8.492 / (0/0169)

= 502(3sf)

7.) Next work ot the Mr of (NH4)2SO4.FeSO4 using your periodic table

so for N (2*14)

for H (8*1)

for S (2*32.1)

for O (8 * 16)

for Fe(1*55.8)

TOTAL = 284

So minus 284 form the total Mr of the salt to gwt the Mr of the water of crystallisation

502-284 = 218

Mr of 1 molecule of H2O is (1*2)+(1*16)= 18

therefore number of water molecule (n)

...... = 218/18

= 12.11 (2dp)

So value of n is 12!!

Phew! hope that explained it xxx

12 years later this is still extremely helpful !! thank you (on the off chance you see this)

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