can someone please help me with question 3iv https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FTopic-Qs%2FEdexcel%2F05-Formulae-Equations-and-Amounts-of-Substance%2FSet-G%2FEquations%2520and%2520Percentage%2520Yield.pdf

thanks!

thanks!

Gosh that does look horrible

I don't know how far you've gone with this, I'll try lay it out

the usual:

find moles of MnO4-

then use the ratio to find mole of C2O4 ions

As you know from start there's a 0.0235 mol dm–3 solution of the salt, you can find the mole of the salt in 25cm3

Then go find the Mr with an equation using y

I hope this helps (I'm halfway confusing myself through this)

I don't know how far you've gone with this, I'll try lay it out

the usual:

find moles of MnO4-

then use the ratio to find mole of C2O4 ions

As you know from start there's a 0.0235 mol dm–3 solution of the salt, you can find the mole of the salt in 25cm3

Then go find the Mr with an equation using y

I hope this helps (I'm halfway confusing myself through this)

(edited 11 months ago)

I'm not entirely sure how to do this either (please let me know if you have the markscheme!!) but I tried working it out, and here's what I did.

So the question states to find out the value of y in the formula KH3(C2O4)y.zH2O (we can ignore the hydrated part because that's in the next question)

They've also given us the equation of the titration with Potassium Permanganate.

2MnO4- + 5C2O4(2-) + 16H+ -> 2Mn(2+) + 10CO2 + 8H2O

From the question, we know that the concentration of MnO4(-) is 0.0203mol/dm^3

The titre of MnO4(-) given to us is 23.15cm^3 --> 0.02315dm^3

Thus we can work out the moles of MnO4(-), which is 0.0203 * 0.02315 = 4.69945*10^(-4) moles

We're also given the concentration of the salt solution (containing C2O4(2-)) which is 0.0235mol/dm^3 and the volume of the solution used to titrate with the manganate, which is 25cm^3 --> 0.025dm^3

If we calculate the moles from the concentration and volume we get 0.0235 * 0.025 = 5.875*10^(-4) moles

If you notice, the MnO4 and C2O4 are in a 2:5 molar ratio but their mole values are not.

If you divide the moles of MnO4 by 2 (as it is 2MnO4(-) in the equation) you get 2.35*10^(-4) moles

then times by 5 to get moles of C2O4 (by molar ratio of 2:5) to get 1.175*10(-3) moles

lastly, if you divide the moles of C2O4 you calculated now by the moles of C2O4 we calculated earlier, you get (1.175*10(-3) / 5.875*10(-4) = 2.

therefore the value of y must be 2

you can cross check this because the charges on the original salt match up.

KH3(C2O4)2

KH3 formed 4+ and C2O4 is a 2- ion compound so if y=2, then (C2O4)2 is -4 and they balance out.

Hope that helped!! (and hopefully it is right otherwise I'm just purely embarrassed for writing all of this)

So the question states to find out the value of y in the formula KH3(C2O4)y.zH2O (we can ignore the hydrated part because that's in the next question)

They've also given us the equation of the titration with Potassium Permanganate.

2MnO4- + 5C2O4(2-) + 16H+ -> 2Mn(2+) + 10CO2 + 8H2O

From the question, we know that the concentration of MnO4(-) is 0.0203mol/dm^3

The titre of MnO4(-) given to us is 23.15cm^3 --> 0.02315dm^3

Thus we can work out the moles of MnO4(-), which is 0.0203 * 0.02315 = 4.69945*10^(-4) moles

We're also given the concentration of the salt solution (containing C2O4(2-)) which is 0.0235mol/dm^3 and the volume of the solution used to titrate with the manganate, which is 25cm^3 --> 0.025dm^3

If we calculate the moles from the concentration and volume we get 0.0235 * 0.025 = 5.875*10^(-4) moles

If you notice, the MnO4 and C2O4 are in a 2:5 molar ratio but their mole values are not.

If you divide the moles of MnO4 by 2 (as it is 2MnO4(-) in the equation) you get 2.35*10^(-4) moles

then times by 5 to get moles of C2O4 (by molar ratio of 2:5) to get 1.175*10(-3) moles

lastly, if you divide the moles of C2O4 you calculated now by the moles of C2O4 we calculated earlier, you get (1.175*10(-3) / 5.875*10(-4) = 2.

therefore the value of y must be 2

you can cross check this because the charges on the original salt match up.

KH3(C2O4)2

KH3 formed 4+ and C2O4 is a 2- ion compound so if y=2, then (C2O4)2 is -4 and they balance out.

Hope that helped!! (and hopefully it is right otherwise I'm just purely embarrassed for writing all of this)

(edited 11 months ago)

Original post by Bookworm524

Gosh that does look horrible

I don't know how far you've gone with this, I'll try lay it out

the usual:

find moles of MnO4-

then use the ratio to find mole of C2O4 ions

As you know from start there's a 0.0235 mol dm–3 solution of the salt, you can find the mole of the salt in 25cm3

Then go find the Mr with an equation using y

I hope this helps (I'm halfway confusing myself through this)

I don't know how far you've gone with this, I'll try lay it out

the usual:

find moles of MnO4-

then use the ratio to find mole of C2O4 ions

As you know from start there's a 0.0235 mol dm–3 solution of the salt, you can find the mole of the salt in 25cm3

Then go find the Mr with an equation using y

I hope this helps (I'm halfway confusing myself through this)

Thank you so much!

This is definitely a hard question

Original post by thechemimech

I'm not entirely sure how to do this either (please let me know if you have the markscheme!!) but I tried working it out, and here's what I did.

So the question states to find out the value of y in the formula KH3(C2O4)y.zH2O (we can ignore the hydrated part because that's in the next question)

They've also given us the equation of the titration with Potassium Permanganate.

2MnO4- + 5C2O4(2-) + 16H+ -> 2Mn(2+) + 10CO2 + 8H2O

From the question, we know that the concentration of MnO4(-) is 0.0203mol/dm^3

The titre of MnO4(-) given to us is 23.15cm^3 --> 0.02315dm^3

Thus we can work out the moles of MnO4(-), which is 0.0203 * 0.02315 = 4.69945*10^(-4) moles

We're also given the concentration of the salt solution (containing C2O4(2-)) which is 0.0235mol/dm^3 and the volume of the solution used to titrate with the manganate, which is 25cm^3 --> 0.025dm^3

If we calculate the moles from the concentration and volume we get 0.0235 * 0.025 = 5.875*10^(-4) moles

If you notice, the MnO4 and C2O4 are in a 2:5 molar ratio but their mole values are not.

If you divide the moles of MnO4 by 2 (as it is 2MnO4(-) in the equation) you get 2.35*10^(-4) moles

then times by 5 to get moles of C2O4 (by molar ratio of 2:5) to get 1.175*10(-3) moles

lastly, if you divide the moles of C2O4 you calculated now by the moles of C2O4 we calculated earlier, you get (1.175*10(-3) / 5.875*10(-4) = 2.

therefore the value of y must be 3

you can cross check this because the charges on the original salt match up.

KH3(C2O4)2

KH3 formed 4+ and C2O4 is a 2- ion compound so if y=2, then (C2O4)2 is -4 and they balance out.

Hope that helped!! (and hopefully it is right otherwise I'm just purely embarrassed for writing all of this)

So the question states to find out the value of y in the formula KH3(C2O4)y.zH2O (we can ignore the hydrated part because that's in the next question)

They've also given us the equation of the titration with Potassium Permanganate.

2MnO4- + 5C2O4(2-) + 16H+ -> 2Mn(2+) + 10CO2 + 8H2O

From the question, we know that the concentration of MnO4(-) is 0.0203mol/dm^3

The titre of MnO4(-) given to us is 23.15cm^3 --> 0.02315dm^3

Thus we can work out the moles of MnO4(-), which is 0.0203 * 0.02315 = 4.69945*10^(-4) moles

We're also given the concentration of the salt solution (containing C2O4(2-)) which is 0.0235mol/dm^3 and the volume of the solution used to titrate with the manganate, which is 25cm^3 --> 0.025dm^3

If we calculate the moles from the concentration and volume we get 0.0235 * 0.025 = 5.875*10^(-4) moles

If you notice, the MnO4 and C2O4 are in a 2:5 molar ratio but their mole values are not.

If you divide the moles of MnO4 by 2 (as it is 2MnO4(-) in the equation) you get 2.35*10^(-4) moles

then times by 5 to get moles of C2O4 (by molar ratio of 2:5) to get 1.175*10(-3) moles

lastly, if you divide the moles of C2O4 you calculated now by the moles of C2O4 we calculated earlier, you get (1.175*10(-3) / 5.875*10(-4) = 2.

therefore the value of y must be 3

you can cross check this because the charges on the original salt match up.

KH3(C2O4)2

KH3 formed 4+ and C2O4 is a 2- ion compound so if y=2, then (C2O4)2 is -4 and they balance out.

Hope that helped!! (and hopefully it is right otherwise I'm just purely embarrassed for writing all of this)

the mark scheme is all the way down the page just scroll down. The answer is 2 but u have done a much better job than me i didnt even attempt it LOOL. thank you so much

Original post by tasneem.016

the mark scheme is all the way down the page just scroll down. The answer is 2 but u have done a much better job than me i didnt even attempt it LOOL. thank you so much

OMG REALLY?? (I was prepared to be humbled after writing a whole long ass explanation with such confidence if it actually turned out to be wrong LOL) but I hope it helped! (I realised I accidentally wrote 3 after calculating it as 2 lmfao) What I find helps me a lot with questions (titration questions that look unnecessarily hard) like these is when you see the word titration, write out the entire equation and underneath each reactant, write down the concentration and the volume (of whichever applies) and then work out the moles. And watch out for molar ratios!! after you work out the moles, it should ideally explain itself!!

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