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 + # H2SO4 +10e^- --> 2 MnO4 + 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.
These questions really scare me!!! I have just started this new chapter in my chemistry lessons and I don't get it at all!!!PLEASE HELP!!
You need to re-read the chemistry book chapter(s). I'm not sure I buy the equations you have written; the first one shows Fe changing from +2 to +3, which is OK, but the second one shows no change in Mn. It should change from +7 on the left to +2 on the right and a more likely product is MnSO4 rather than MnO4. The second point is to remember in these reactions that
equivalent mass = molar mass/delta e per mole. For FeSO4 that is molar mass/1. For KMnO4 that is molar mass/5.
For #3, normality = # equivalents/liter of solution.
For #4, use mL x N = mL x N
For #5, this is a standard redox equation to balance; redox equations are balanced following rules. Here is a site that give some simple rules.
http://www.chemteam.info/Redox/Redox.html
Don't worry, I'm here to help you understand step by step. Let's go through each question one by one.
1. To find the gram-equivalent mass of FeSO4, we need to consider the number of moles of FeSO4 that are involved in the reaction. From the balanced half-reaction, we see that 10 moles of FeSO4 produce 10 moles of electrons. Since the molar mass of FeSO4 is 151.91 g/mol, the gram-equivalent mass of FeSO4 is (151.91 g/mol) / 10 = 15.19 g/equivalent.
2. Similarly, for KMnO4, the balanced half-reaction shows that 10 moles of electrons are involved. The molar mass of KMnO4 is 158.04 g/mol, so the gram-equivalent mass is (158.04 g/mol) / 10 = 15.804 g/equivalent.
3. The normality of a solution is the number of gram-equivalents of solute per liter of solution. In this case, we are given that 5.35 g of KMnO4 are dissolved in a liter of solution. From question 2, we know that the gram-equivalent mass of KMnO4 is 15.804 g/equivalent. Therefore, the normality of the solution is (5.35 g) / (15.804 g/equivalent) = 0.339 N.
4. Now let's tackle the titration problem. We are given that 65.34 mL of 0.1000 N KMnO4 are required to titrate 45.28 mL of FeSO4 solution. From the balanced equation, we can deduce that the stoichiometric ratio between KMnO4 and FeSO4 is 2:10 (from the coefficients in the balanced half-reactions). Therefore, we can set up a proportion:
(65.34 mL KMnO4) / (45.28 mL FeSO4) = (0.1000 N KMnO4) / (x N FeSO4)
Solving for x, the normality of FeSO4, we get x = (65.34 mL KMnO4) / (45.28 mL FeSO4) * (0.1000 N KMnO4) = 0.143 N.
5. Lastly, let's balance the chemical equation given. The unbalanced equation is:
CO + I2O5 --> CO2 + I2
To balance it, we start by counting the number of atoms on each side. We have 1 C, 1 O, 2 I on the left and 1 C, 3 O, 2 I on the right.
By adjusting the coefficients, we can balance the equation as follows:
CO + 2 I2O5 --> CO2 + 5 I2
And that's it! I hope this step-by-step explanation helps you understand the concepts better. If you have any further questions, feel free to ask.
I'll do my best to help you with these questions! Chemistry can be challenging, but with some guidance, you'll be able to understand these concepts. Let's take it step by step:
1. The gram-equivalent mass of a substance is the mass of that substance that combines with or replaces one mole of hydrogen ions (H^+). Since the balanced equation indicates that 10 moles of FeSO4 are required to react with 10 moles of H^+, we can calculate the gram-equivalent mass of FeSO4 as follows:
Gram-equivalent mass of FeSO4 = Molar mass of FeSO4 / 10
Given that the molar mass of FeSO4 is 151.91 g/mol, we can find the gram-equivalent mass:
Gram-equivalent mass of FeSO4 = 151.91 g/mol / 10 = 15.191 g
2. Similar to the previous question, the gram-equivalent mass of KMnO4 can be calculated using the balanced equation. From the equation, we can see that 10 moles of electrons (e^-) are required for 2 moles of KMnO4. Therefore, the gram-equivalent mass of KMnO4 is:
Gram-equivalent mass of KMnO4 = Molar mass of KMnO4 / 10
Given that the molar mass of KMnO4 is 158.04 g/mol, we can find the gram-equivalent mass:
Gram-equivalent mass of KMnO4 = 158.04 g/mol / 10 = 15.804 g
3. Normality is a measure of the concentration of a solution. It is defined as the number of equivalents of a solute in one liter of solution. To calculate the normality of the KMnO4 solution, we need to know the number of equivalents of KMnO4 present.
To find the number of equivalents, we need to know the molarity of KMnO4. Unfortunately, it's not provided in the question. Therefore, we cannot determine the normality with the information given.
4. In this question, we are given the volume (in mL) and normality of KMnO4 used in a titration, as well as the volume (in mL) of FeSO4 solution. To find the normality of the FeSO4 solution, we can use the following formula:
Normality of FeSO4 = Normality of KMnO4 * Volume of KMnO4 / Volume of FeSO4
Given that the normality of KMnO4 is 0.1000 N, the volume of KMnO4 used is 65.34 mL, and the volume of FeSO4 solution is 45.28 mL, we can substitute these values into the formula:
Normality of FeSO4 = 0.1000 N * 65.34 mL / 45.28 mL
Solving this equation will give you the normality of the FeSO4 solution.
5. The balanced equation you provided is already balanced:
CO + I2O5 --> CO2 + I2
No further balancing is necessary.
I hope this helps! Remember to always pay attention to the units and the balanced equations when solving chemistry problems. If you have any more questions, feel free to ask!