So I posted this question yesterday and was told I had a few typos.... I fixed them and here is the question again... Please help me!!!

A solution of ferrous sulfate (FeSO4) is titrated with a solution of potassium permanganate in sulfuric acid. The balanced half reactions are as follows:

10FeSO4 + 5 H2SO4 --> 5 Fe2(SO4)3 + 10 H ^+ +10e^-
and 2 KMnO4 + 3 H2SO4 +10e^- --> 2 MnSO4 + K2SO4 + 8 H2O.

The gram-formula mass of FeSO4 is 151.91 g/mol, and that of KMnO4 is 158.04 g/mol.

1. What is the gram-equavalent mass of FeSo4 for this reaction?

2. What is the gram-equavalent mass of KMnO4 for this reaction?

3. If 5.35 g of KMnO4 are dissolved in
a liter of solution, what is the normality of the solution?

4. If 65.34 mL of 0.1000 N KMnO4 are required to titrate 45.28 mL of FeSO4 solution, what is the normality of the solution?

5. Carbon monoxide is detected in air samples by its reaction with diiodine pentoxide to form carbon dioxide and elemental iodine: CO + I2O5 --> CO2 + I2. Balance the equation.

my answers so far...

1. 151.91g FeSO4/1eq

2. 31.608g KMnO4/1eq

3. N=5eq/1L (not sure about this one)

Sure, I can help you with that! Let's go through each question step by step:

1. To find the gram-equivalent mass of FeSO4, we need to know the molar mass of FeSO4. The molar mass of FeSO4 is given as 151.91 g/mol. The gram-equivalent mass is obtained by dividing the molar mass by the number of equivalents.

Since the balanced half-reaction shows that FeSO4 donates 10 electrons, we can also say that one mole of FeSO4 donates 10 equivalents of electrons. Thus, the gram-equivalent mass of FeSO4 is calculated as:

Gram-equivalent mass = Molar mass / Number of equivalents
= 151.91 g/mol / 10 equiv
= 15.191 g/equiv

Therefore, the gram-equivalent mass of FeSO4 for this reaction is 15.191 g/equiv.

2. Similarly, to find the gram-equivalent mass of KMnO4, we need to know the molar mass of KMnO4. The molar mass of KMnO4 is given as 158.04 g/mol. In the balanced half-reaction, we see that KMnO4 accepts 10 electrons, so it donates 10 equivalents.

Gram-equivalent mass of KMnO4 = Molar mass / Number of equivalents
= 158.04 g/mol / 10 equiv
= 15.804 g/equiv

Therefore, the gram-equivalent mass of KMnO4 for this reaction is 15.804 g/equiv.

3. The normality of a solution is defined as the number of equivalents of solute per liter of solution. In this case, 5.35 g of KMnO4 is dissolved in 1 L of solution. To find the normality of the solution, we need to calculate the number of equivalents of KMnO4.

Number of equivalents = Mass of solute / Gram-equivalent mass
= 5.35 g / 15.804 g/equiv
≈ 0.339 equiv

Therefore, the normality of the KMnO4 solution is approximately 0.339 N.

4. In this question, we have a titration where 65.34 mL of 0.1000 N KMnO4 is required to titrate 45.28 mL of FeSO4 solution. We need to find the normality of the FeSO4 solution.

We can use the equation: Normality of KMnO4 × Volume of KMnO4 solution = Normality of FeSO4 × Volume of FeSO4 solution

Normality of FeSO4 = (Normality of KMnO4 × Volume of KMnO4 solution) / Volume of FeSO4 solution

Plugging in the values: Normality of FeSO4 = (0.1000 N × 65.34 mL) / 45.28 mL

Therefore, the normality of the FeSO4 solution is approximately 0.145 N.

5. To balance the equation CO + I2O5 → CO2 + I2, we need an equal number of carbon, oxygen, and iodine atoms on both sides.

The balanced equation would be: 2CO + I2O5 → 2CO2 + I2

In this balanced equation, we have two carbon atoms, five oxygen atoms, and two iodine atoms on each side. Therefore, the equation is now balanced.

I hope this helps! Let me know if you have any further questions.