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Moles help
FA 1is a solution containing 5.00 g dm–3 of hydrated ethanedioic acid, H2C2O4.xH2O.
FA 2 is a solution containing 2.37 g dm–3 of potassium manganate(VII), KMnO4.
You are also provided with 1.00moldm–3 sulphuric acid, H2SO4.
In the presence of acid, potassium manganate(VII) oxidises ethanedioic acid;
2(MnO4)– (aq) + 5H2C2O4(aq) + 6H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
You are to determine the value of x in H2C2O4.xH2O.
My titre (amount of FA2 used was) 26.58cm^2
(b) Calculate how many moles of potassium manganate(VII), KMnO4, were run from the
burette during the titration.
[Ar: K, 39.1; Mn, 54.9; O, 16.0.]
stuck
(c) Calculate how many moles of ethanedioic acid, H2C2O4, reacted with the potassium
manganate(VII) run from the burette.
my answer in part(b) x 5/2
(d) Calculate the mass of H2C2O4 in each dm3 of FA 1
[Ar: H, 1.0; C, 12.0; O, 16.0.]
stuck
(e) Calculate the mass of water in the 5.00 g of H2C2O4.xH2O.
stuck
(f) Calculate the value of x, in H2C2O4.xH2O. [1]
stuck
FA 2 is a solution containing 2.37 g dm–3 of potassium manganate(VII), KMnO4.
You are also provided with 1.00moldm–3 sulphuric acid, H2SO4.
In the presence of acid, potassium manganate(VII) oxidises ethanedioic acid;
2(MnO4)– (aq) + 5H2C2O4(aq) + 6H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
You are to determine the value of x in H2C2O4.xH2O.
My titre (amount of FA2 used was) 26.58cm^2
(b) Calculate how many moles of potassium manganate(VII), KMnO4, were run from the
burette during the titration.
[Ar: K, 39.1; Mn, 54.9; O, 16.0.]
stuck
(c) Calculate how many moles of ethanedioic acid, H2C2O4, reacted with the potassium
manganate(VII) run from the burette.
my answer in part(b) x 5/2
(d) Calculate the mass of H2C2O4 in each dm3 of FA 1
[Ar: H, 1.0; C, 12.0; O, 16.0.]
stuck
(e) Calculate the mass of water in the 5.00 g of H2C2O4.xH2O.
stuck
(f) Calculate the value of x, in H2C2O4.xH2O. [1]
stuck
arsenalfan.
FA 1is a solution containing 5.00 g dm–3 of hydrated ethanedioic acid, H2C2O4.xH2O.
FA 2 is a solution containing 2.37 g dm–3 of potassium manganate(VII), KMnO4.
You are also provided with 1.00moldm–3 sulphuric acid, H2SO4.
In the presence of acid, potassium manganate(VII) oxidises ethanedioic acid;
2(MnO4)– (aq) + 5H2C2O4(aq) + 6H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
You are to determine the value of x in H2C2O4.xH2O.
My titre (amount of FA2 used was) 26.58cm^2
(b) Calculate how many moles of potassium manganate(VII), KMnO4, were run from the
burette during the titration.
[Ar: K, 39.1; Mn, 54.9; O, 16.0.]
FA 2 is a solution containing 2.37 g dm–3 of potassium manganate(VII), KMnO4.
You are also provided with 1.00moldm–3 sulphuric acid, H2SO4.
In the presence of acid, potassium manganate(VII) oxidises ethanedioic acid;
2(MnO4)– (aq) + 5H2C2O4(aq) + 6H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
You are to determine the value of x in H2C2O4.xH2O.
My titre (amount of FA2 used was) 26.58cm^2
(b) Calculate how many moles of potassium manganate(VII), KMnO4, were run from the
burette during the titration.
[Ar: K, 39.1; Mn, 54.9; O, 16.0.]
The solution strength, 2.37 g dm–3, multiplied by the number of litres that you used gives you the number of grams.
then use moles = mass/RMM to get the number of moles of potassium manganate VII
arsenalfan.
(c) Calculate how many moles of ethanedioic acid, H2C2O4, reacted with the potassium
manganate(VII) run from the burette.
my answer in part(b) x 5/2
manganate(VII) run from the burette.
my answer in part(b) x 5/2
ok
arsenalfan.
(d) Calculate the mass of H2C2O4 in each dm3 of FA 1
[Ar: H, 1.0; C, 12.0; O, 16.0.]
(d) Calculate the mass of H2C2O4 in each dm3 of FA 1
[Ar: H, 1.0; C, 12.0; O, 16.0.]
If you have the moles of H2C2O4 find the mass from mass = moles x RMM
arsenalfan.
(e) Calculate the mass of water in the 5.00 g of H2C2O4.xH2O.
(e) Calculate the mass of water in the 5.00 g of H2C2O4.xH2O.
if you have the mass of H2C2O4 subtract this from the 5g use to find the mass of water
arsenalfan.
(f) Calculate the value of x, in H2C2O4.xH2O. [1]
(f) Calculate the value of x, in H2C2O4.xH2O. [1]
Now divide by 18 to get the moles of water.
You now have the moles of H2C2O4 and the moles of water
Find the simplest ratio by dividing both values by the moles of H2C2O4
chatchi
Thanks for your help!
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