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nx01rules

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Question 1. An 8.75g. sample of iron wire was dissolved in excess sulfuric acid. All the iron present formed iron (II) ions. This solution was made up to 1.00 L. with distilled water. A 20.0 mL. sample of this solution was pipetted into a conical flask and titrated with 0.025 mol/L potassium permanganate solution. An average titre value of 22.1 mL. of permanganate was required to reach the equivalence point

a) Calculate (showing all the working) the concentration of iron (II) ions in the 1.00 L. solution if the equation

for the reaction was

5Fe2+(aq) + 2MnO4-(aq) + 16H+(aq) = 5Fe3+(aq) + 2Mn2+(aq) + 8H2O(l)

b) Calculate the % by mass of iron in the original 8.75g sample of wire.

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I am thinking that you need to work out the number of mols of Iron (II) present, using n=m/M, and then use n=CV, and rearrange it into C=n/V and work out the concentration. But I also have a nagging feeling that you need to take into account the mol ratios and use something like n(Fe)/5 = n(MnO4)/2. But I can't use n(Fe), because it is a solid and I need to use m/M for it, and it doesn't have concentration in that equation, which is what I am trying to work out.

Please, some assistance would be fantastic!

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Question 2. What volume of 0.125 mol/L H2SO4 must be added to 1.25 L of 0.10 mol/L HCl in order to exactly dissolve

4.5g of Magnesium? (Show all the working out.)

The equations for the reactions are:

2HCl + Mg = MgCl2 + H2

AND:

Mg + H2SO4 = MgSO4 + H2

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I tried to work out Question 2 after I wrote this, I think I may have it right. What I did, was I worked out the number of mols of HCl present:

(I assume you are familiar with n=m/M and n=CV )

n(hcl)/2 = n(mg)/1 (The division bits are dividing each side by their mole ratio, obtained from the equation)

Then I began to work out exactly how much Magnesium would actually react with that certain number of mols of HCl:

(C X V)/2 = m/M (Where m is the mass in grams, M is the molar mass in g/mol)

Filling in all the known values:

(0.10 X 1.25)/2 = m/24.31

After rearranging the equation to get m, I came up with:

(0.125 X 24.31)/2 = m

Therefore, m = 1.52g

So, I found that 1.52g of magnesium will react with the HCl. Now, what is remaining will react with the H2SO4. The

amount remaining is:

4.5g-1.52g = 2.98g

Then, I worked out how much H2SO4 is needed using another stoichiometric calculation:

n(H2SO4) = n(Mg)

C X V = m/M

The remaining Magnesium is used as well to help calculate how much H2SO4 is needed to neutralise what is left of it:

0.125 X V = 2.98/24.31

After rearranging:

V = 0.123/0.125

V = 0.98L

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I think this question would be all good and right, if not, please correct me. If it is right, can I please have some

help on Question 1?

Thanks to anyone who can help!