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3 years ago

hi , not sure where to post this , but i hope it reaches someone .

A 1.0101 gram sample of Fe(OH)n was mixed with 20.00 mL of 2.000 M HCl and enough water added to make 200.0 mL of solution A.

Fe(OH)n(s) + nHCl(aq) FeCln(aq) + nH2O(l)

A 25.00 mL aliquot of solution A was taken and titrated with 14.56 mL of 0.1000 M KOH.

KOH(aq) + HCl(aq) H2O(l) + KCl(aq)

What is the value of n?

i found this answer online :

0.01456 L * 0.1000 mol/L * 200/25 = 0.011648 moles HCl that didn't react.

0.020 L * 2.000 mol/L = 0.0400 original mol HCl.

0.0400 - 0.011648 = 0.028352 moles HCl reacted.

Assuming n = 3, then 1.0101 g / 106.867 g/mol =0.00945 mol.

Ratio = 0.02845 / 0.00945 = 3.00.

I had to assume n was 3 to get the molar mass to make sense of the mass.

The Fe(OH)n / HCl ratio was 3 based on this assumption.

the answer is 3 , i don’t understand why or how we assume that n=3 , or the 1.0101/ 106.867 calculation . Where did the last number come from ? thank you

A 1.0101 gram sample of Fe(OH)n was mixed with 20.00 mL of 2.000 M HCl and enough water added to make 200.0 mL of solution A.

Fe(OH)n(s) + nHCl(aq) FeCln(aq) + nH2O(l)

A 25.00 mL aliquot of solution A was taken and titrated with 14.56 mL of 0.1000 M KOH.

KOH(aq) + HCl(aq) H2O(l) + KCl(aq)

What is the value of n?

i found this answer online :

0.01456 L * 0.1000 mol/L * 200/25 = 0.011648 moles HCl that didn't react.

0.020 L * 2.000 mol/L = 0.0400 original mol HCl.

0.0400 - 0.011648 = 0.028352 moles HCl reacted.

Assuming n = 3, then 1.0101 g / 106.867 g/mol =0.00945 mol.

Ratio = 0.02845 / 0.00945 = 3.00.

I had to assume n was 3 to get the molar mass to make sense of the mass.

The Fe(OH)n / HCl ratio was 3 based on this assumption.

the answer is 3 , i don’t understand why or how we assume that n=3 , or the 1.0101/ 106.867 calculation . Where did the last number come from ? thank you

Reply 1

3 years ago

Original post by sierranonis

hi , not sure where to post this , but i hope it reaches someone .

the answer is 3 , i don’t understand why or how we assume that n=3 , or the 1.0101/ 106.867 calculation . Where did the last number come from ? thank you

the answer is 3 , i don’t understand why or how we assume that n=3 , or the 1.0101/ 106.867 calculation . Where did the last number come from ? thank you

Hi

They've used trial and error to work it out, they know that in all likelihood that n = 2 or 3.

However, you can work it out analytically:

Moles of HCl reacted is 0.028352 mol, but this reacts with Fe(OH)n in a ratio of 1:n so therefore moles of Fe(OH)n = 0.028352/n

Then equating this to moles = mass/Mr gives us 0.028352/n = 1.0101/(56+17n)

which then solves for n = 3

Reply 2

3 years ago

Original post by EierVonSatan

Hi

They've used trial and error to work it out, they know that in all likelihood that n = 2 or 3.

However, you can work it out analytically:

Moles of HCl reacted is 0.028352 mol, but this reacts with Fe(OH)n in a ratio of 1:n so therefore moles of Fe(OH)n = 0.028352/n

Then equating this to moles = mass/Mr gives us 0.028352/n = 1.0101/(56+17n)

which then solves for n = 3

They've used trial and error to work it out, they know that in all likelihood that n = 2 or 3.

However, you can work it out analytically:

Moles of HCl reacted is 0.028352 mol, but this reacts with Fe(OH)n in a ratio of 1:n so therefore moles of Fe(OH)n = 0.028352/n

Then equating this to moles = mass/Mr gives us 0.028352/n = 1.0101/(56+17n)

which then solves for n = 3

Thank you so much !

but could u explain why and where the numbers came from ? in terms of 1.0101 . i’m so confused as to what i’m even calculating ahhh thank you

Reply 3

3 years ago

Original post by sierranonis

Thank you so much !

but could u explain why and where the numbers came from ? in terms of 1.0101 . i’m so confused as to what i’m even calculating ahhh thank you

but could u explain why and where the numbers came from ? in terms of 1.0101 . i’m so confused as to what i’m even calculating ahhh thank you

1.0101g is the mass of the solid in the question

Sorry, I assumed that you understood the working you posted up to the point where you asked about where n = 3 came from.

moles of HCl added = 0.04 mol to make solution A

moles of KOH added for 25cm3 titration = 1.456... x 10

moles of KOH needed for 200cm3 = 8 x 1.456... x 10

moles of HCl reacted before the titration, making up solution A = 0.04 - 0.011648 = 0.028352 mol

Mr = 56 + 17n (from Fe and n x OH)

Reply 4

3 years ago

Original post by EierVonSatan

Hi

They've used trial and error to work it out, they know that in all likelihood that n = 2 or 3.

However, you can work it out analytically:

Moles of HCl reacted is 0.028352 mol, but this reacts with Fe(OH)n in a ratio of 1:n so therefore moles of Fe(OH)n = 0.028352/n

Then equating this to moles = mass/Mr gives us 0.028352/n = 1.0101/(56+17n)

which then solves for n = 3

They've used trial and error to work it out, they know that in all likelihood that n = 2 or 3.

However, you can work it out analytically:

Moles of HCl reacted is 0.028352 mol, but this reacts with Fe(OH)n in a ratio of 1:n so therefore moles of Fe(OH)n = 0.028352/n

Then equating this to moles = mass/Mr gives us 0.028352/n = 1.0101/(56+17n)

which then solves for n = 3

Thank you so much !

but could u explain why and where the numbers came from ? in terms of 1.0101 . i’m so confused as to what i’m even calculating ahhh thank you

Original post by EierVonSatan

1.0101g is the mass of the solid in the question

Sorry, I assumed that you understood the working you posted up to the point where you asked about where n = 3 came from.

moles of HCl added = 0.04 mol to make solution A

moles of KOH added for 25cm3 titration = 1.456... x 10^{-3} mol

moles of KOH needed for 200cm3 = 8 x 1.456... x 10^{-3} mol = 0.011648 mol

moles of HCl reacted before the titration, making up solution A = 0.04 - 0.011648 = 0.028352 mol

Mr = 56 + 17n (from Fe and n x OH)

Sorry, I assumed that you understood the working you posted up to the point where you asked about where n = 3 came from.

moles of HCl added = 0.04 mol to make solution A

moles of KOH added for 25cm3 titration = 1.456... x 10

moles of KOH needed for 200cm3 = 8 x 1.456... x 10

moles of HCl reacted before the titration, making up solution A = 0.04 - 0.011648 = 0.028352 mol

Mr = 56 + 17n (from Fe and n x OH)

THANK YOU SM AHHH😭😭 i’ve spent hours on this 😭 and wow , my brain is so tired , i didn’t even see the mass number in the question ahh! thank you sm though , have a good nightttt

Reply 5

3 years ago

Original post by EierVonSatan

1.0101g is the mass of the solid in the question

Sorry, I assumed that you understood the working you posted up to the point where you asked about where n = 3 came from.

moles of HCl added = 0.04 mol to make solution A

moles of KOH added for 25cm3 titration = 1.456... x 10^{-3} mol

moles of KOH needed for 200cm3 = 8 x 1.456... x 10^{-3} mol = 0.011648 mol

moles of HCl reacted before the titration, making up solution A = 0.04 - 0.011648 = 0.028352 mol

Mr = 56 + 17n (from Fe and n x OH)

Sorry, I assumed that you understood the working you posted up to the point where you asked about where n = 3 came from.

moles of HCl added = 0.04 mol to make solution A

moles of KOH added for 25cm3 titration = 1.456... x 10

moles of KOH needed for 200cm3 = 8 x 1.456... x 10

moles of HCl reacted before the titration, making up solution A = 0.04 - 0.011648 = 0.028352 mol

Mr = 56 + 17n (from Fe and n x OH)

Hey , it's a week later and im redoing this question , could you please explain why KOH is calculated for 200cm3 ? Im so confused ahhh. Also , where did the HCL in the second equation even come from?

(edited 3 years ago)

Reply 6

3 years ago

Original post by sierranonis

Hey , it's a week later and im redoing this question , could you please explain why KOH is calculated for 200cm3 ? Im so confused ahhh. Also , where did the HCL in the second equation even come from?

Because only a small sample was titrated with KOH, you need amount needed to react with the mass of solid that was dissolved in 200cm3.

As for the acid: "was mixed with 20.00 mL of 2.000 M HCl"

Reply 7

2 years ago

Original post by EierVonSatan

Hi

They've used trial and error to work it out, they know that in all likelihood that n = 2 or 3.

However, you can work it out analytically:

Moles of HCl reacted is 0.028352 mol, but this reacts with Fe(OH)n in a ratio of 1:n so therefore moles of Fe(OH)n = 0.028352/n

Then equating this to moles = mass/Mr gives us 0.028352/n = 1.0101/(56+17n)

which then solves for n = 3

They've used trial and error to work it out, they know that in all likelihood that n = 2 or 3.

However, you can work it out analytically:

Moles of HCl reacted is 0.028352 mol, but this reacts with Fe(OH)n in a ratio of 1:n so therefore moles of Fe(OH)n = 0.028352/n

Then equating this to moles = mass/Mr gives us 0.028352/n = 1.0101/(56+17n)

which then solves for n = 3

Hi again,

I'm going over this question again, and I see that you said you assumed that it was 2 or 3. What are we using 2 or 3 to test for to determine whether it is 2 or 3 though? How do we test if the value is 2 or 3? I hope that makes sense.

Reply 8

2 years ago

Original post by sierranonis

Hi again,

I'm going over this question again, and I see that you said you assumed that it was 2 or 3. What are we using 2 or 3 to test for to determine whether it is 2 or 3 though? How do we test if the value is 2 or 3? I hope that makes sense.

I'm going over this question again, and I see that you said you assumed that it was 2 or 3. What are we using 2 or 3 to test for to determine whether it is 2 or 3 though? How do we test if the value is 2 or 3? I hope that makes sense.

I hope you haven't been stuck on this all this time

I didn't assume - that they did The common charges of iron ions are 2+ and 3+

Reply 9

2 years ago

Original post by EierVonSatan

I hope you haven't been stuck on this all this time

I didn't assume - that they did The common charges of iron ions are 2+ and 3+

I didn't assume - that they did The common charges of iron ions are 2+ and 3+

Ahh, I genuinely thought I understood it last time (a few months ago), and came back to it again to recap and I'm confusedddd. I got that part (Fe 2+ or 3+), but I mean how do you use either value to figure out which value it is? As in what do you do with the numbers to find out which one is N?

They've used trial and error to work it out, they know that in all likelihood that n = 2 or 3.

However, you can work it out analytically:

Moles of HCl reacted is 0.028352 mol, but this reacts with Fe(OH)n in a ratio of 1:n so therefore moles of Fe(OH)n = 0.028352/n

Then equating this to moles = mass/Mr gives us 0.028352/n = 1.0101/(56+17n)

which then solves for n = 3

Hiya, doing this question right now. How come the moles of Fe(OH)n = 0.028352/n. That's the only thing I can't manage to wrap my head around. Thanks!

(edited 2 months ago)

Reply 11

2 months ago

Original post by Azdin

Hiya, doing this question right now. How come the moles of Fe(OH)n = 0.028352/n. That's the only thing I can't manage to wrap my head around. Thanks!

Due to the ratio in which they react. Every mole of Fe(OH)n reacts with n moles of acid:

e.g. 2HCl + Fe(OH)2 ---> FeCl2 + 2H2O

3HCl + Fe(OH)3 ---> FeCl3 + 3H2O

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