NaimR
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Hi I've come across a giant moles calculations question in unit 5, I've worked through it but I'm not sure if I have the correct answer cause I can't find the markscheme :confused:

Can anyone help me with the method I'd need to take with this sort of question (I'm not just asking for the answer outright)

The following method was used to determine the percentage by mass of vanadium in a sample of ammonium vanadate(V).
A solution was made up by dissolving 0.160 g of ammonium vanadate(V) in dilute sulphuric acid. The ammonium vanadate(V) formed ions in this solution. When an excess of zinc was added to this solution, the ions were reduced to V2+ ions and the zinc was oxidised to Zn2+ ions.
After the unreacted zinc had been removed, the solution was titrated against a 0.0200 mol dm–3 solution of potassium manganate(VII). In the titration, 38.5 cm3 of potassium manganate(VII) solution were required to oxidise all vanadium(II) ions to vanadium(V) ions.
Using half-equations, construct an overall equation for the reduction of to V2+ by zinc in acidic solution.
Calculate the percentage by mass of vanadium in the sample of ammonium vanadate(V).

(8 marks)
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charco
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(Original post by NaimR)
Hi I've come across a giant moles calculations question in unit 5, I've worked through it but I'm not sure if I have the correct answer cause I can't find the markscheme :confused:

Can anyone help me with the method I'd need to take with this sort of question (I'm not just asking for the answer outright)

The following method was used to determine the percentage by mass of vanadium in a sample of ammonium vanadate(V).
A solution was made up by dissolving 0.160 g of ammonium vanadate(V) in dilute sulphuric acid. The ammonium vanadate(V) formed ions in this solution. When an excess of zinc was added to this solution, the ions were reduced to V2+ ions and the zinc was oxidised to Zn2+ ions.
After the unreacted zinc had been removed, the solution was titrated against a 0.0200 mol dm–3 solution of potassium manganate(VII). In the titration, 38.5 cm3 of potassium manganate(VII) solution were required to oxidise all vanadium(II) ions to vanadium(V) ions.
Using half-equations, construct an overall equation for the reduction of to V2+ by zinc in acidic solution.
Calculate the percentage by mass of vanadium in the sample of ammonium vanadate(V).

(8 marks)
You actually do not need to do any moles calculations to answer the question BUT they want you to so here goes:

Start backwards and calculate the moles of potassium manganate(VII).

Then you know that manganate(VII) turns to manganese(II) so there is a transfer of 5 electrons.

At the same time the vanadate(II) is getting oxidised to vanadate(V), i.e. a transfer of 3 electrons.

So the mole ratio of manganate(VII) to vanadate(II) is 3:5

So multiply the moles of manganate(VII) by 5/3 to get moles of vanadium.

Mol vanadium ==> mass of vanadium

Compare this with initial mass given of ammonium vanadate to find percentage
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NaimR
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(Original post by charco)
You actually do not need to do any moles calculations to answer the question BUT they want you to so here goes:

Start backwards and calculate the moles of potassium manganate(VII).

Then you know that manganate(VII) turns to manganese(II) so there is a transfer of 5 electrons.

At the same time the vanadate(II) is getting oxidised to vanadate(V), i.e. a transfer of 3 electrons.

So the mole ratio of manganate(VII) to vanadate(II) is 3:5

So multiply the moles of manganate(VII) by 5/3 to get moles of vanadium.

Mol vanadium ==> mass of vanadium

Compare this with initial mass given of ammonium vanadate to find percentage
Oh okay thank you I got my percentage by mass as 40.8%, does anyone know if this is correct or has anyone else got the same answer?
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AbdulRaqib
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that's the correct answer!
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adichemist
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(Original post by AbdulRaqib)
that's the correct answer!
wooooo that's what I got! Cheers from 2021!!
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