How Do You Solve Complex Chemistry Titration and Reaction Problems?

In summary, the conversation discusses two chemistry questions, one involving the calculation of the concentration of iron (II) ions in a solution and the other involving the amount of acid needed to dissolve a certain amount of magnesium. The solution to the first question involves using a stoichiometric calculation to find the concentration, while the solution to the second question involves using mole ratios and molar mass to find the percentage of iron in the original sample.
  • #1
nx01rules
5
0
Hi, I am doing chemistry at my school, and I am currently stuck on a couple of questions that I don't know how to complete. Here are the questions:
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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 :biggrin: )

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!
 
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  • #2
An update - it seems i DO have Question 2 right after all - so its just Question 1 that I am in the dark about.
 
  • #3
?

So anyone doesn't know how to do this?
 
  • #4
I could give it a shot. I'm not completely sure I understand what is going on in the problem though.

I'm thinking it could be done backwards. The acid was in excess so just ignore it. For the permanganate, you know the solution concentration and how much it took; find how many moles of permanganate the titration took. Then use the mole ratio to find how much Fe2 was present.

(x moles MnO4) * (5Fe2/2MnO4) = (y moles Fe2)

The concentration for the 1L solution should be the same as the 20mL sample. Divide (y moles Fe2) by 0.02L and you should get the proper concentration.

I don't understand part B.
 
  • #5
Yeah I actually started to think that this problem isn't so hard after all and it doesn't need too much hard thought at all. But for part B, if you work out the number of mols of Fe2+ used IN THE REACTION, you can divide n by Fe's molar mass to work out the MASS of Fe2+ in the solution that was used up. Then, because that will be a smaller value than the original wire, you can find the percentage by mass by going - m/8.75g * 100, where m is the worked out mass. The reason it is less is because the Fe2+ isn't a pure wire. It has impurities in it.
 
  • #6
Q1.

a) 5Fe2+ + 2MnO4- + 16H+ = 5Fe3+ + 2Mn2+ 8H2O

No. of moles of MnO4- in 22.1 L= CxV/1000(since the volume is in mL)
= 22.1 x 0.025/1000
=5.525 x 10^-4 mol

Therefore the no. of moles of Fe2+ reacted with MnO4-=5/2 x 5.525 x 10^-4
(since mole ratio of Fe2+:MnO4-=5:2) =1.38125 x 10^-3mol

This 1.38125 x 10^-3 mol of Fe2+ is in 20 mL of the solution.
Therefore, No. of moles of Fe2+= C x V/1000
1.38125 x 10^-3= C x 20/1000
C= 0.0690625 mol/L

b) Fe + H2SO4 = FeSO4 + H2

In 1L no. of moles of Fe2+ is 0.0690625

Therefore the no. of moles of Fe is also the same ( from the equation the mole ratio of Fe:Fe2+=1:1)=0.0690625 mol

No. of moles=Mass/Molar mass
Mass=No. of moles x Molar mass
=0.0690625 x 56
=3.8675 g

This is the actual mass of Fe present in the 8.75 g of wire.

Therefore % by mass of Fe present=Actual mass of Fe/Mass of wire x 100
=3.8675/8.75 x 100
=44.2%

:blushing: Hope that helped!
 
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FAQ: How Do You Solve Complex Chemistry Titration and Reaction Problems?

1. What are the 2 chemistry questions that you are stuck on?

The 2 chemistry questions that I am stuck on are as follows:

1. What is the difference between an element and a compound?

2. How do you balance a chemical equation?

2. Can you explain the difference between an element and a compound?

An element is a substance that cannot be broken down into simpler substances by chemical means. It is made up of only one type of atom. A compound, on the other hand, is a substance made up of two or more elements that are chemically bonded together. It can be broken down into simpler substances by chemical means.

3. How do you balance a chemical equation?

To balance a chemical equation, you must make sure that the number of atoms of each element on the reactant side is equal to the number of atoms of that element on the product side. This is done by adding coefficients in front of the chemical formulas to adjust the number of each type of atom.

4. What is the purpose of balancing a chemical equation?

The purpose of balancing a chemical equation is to follow the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. By balancing the equation, we ensure that the number of atoms of each element remains the same on both sides of the equation.

5. Can you provide an example of balancing a chemical equation?

Sure, the equation for the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O) is:
2H2 + O2 → 2H2O
To balance this equation, we need to add a coefficient of 2 in front of the hydrogen gas on the reactant side to make the number of hydrogen atoms equal on both sides. This gives us:
2H2 + O2 → 2H2O
Next, we need to add a coefficient of 2 in front of the water on the product side to balance the oxygen atoms. This gives us the balanced equation:
2H2 + O2 → 2H2O

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