A 3.4g sample of h2o2 solution containing x% H2O2 by mass requires x ml of a KMnO4 solution for complete oxidation under acidic condition. the molarity of kmno4 solution is: a)1 b)0.5 c)0.4 d)0.2
assumes the x%(which is in grams) and x mL is the same and they can't be. But I think the following is what the author expects so here is the solution.
2KMnO4 + 5H2O2 + 3H2SO4 =>2MnSO4 + K2SO4 + 8H2O + 5O2
Assuming the x values CAN be the same let's try the 4 answers and see if they fit. For 1 M, we have 1M(xmL) = 1 millimils KMnO4. That will equal (5x/2) mmols H2O2 and (5x/2)*0.034(the millimolar mass H2O2) = 0.085x g H2O2. Then
[(0.085x g)/3.4]*100 = x%. Check it out and the left side doesn't equal the right side.
Do the same for 0.5M, 0.4M and 0.2M
One of them matches. Post your work if you get stuck.
To find the molarity of the KMnO4 solution, we need to use the balanced chemical equation for the reaction between H2O2 and KMnO4 under acidic conditions:
5H2O2 + 2KMnO4 + 6H+ → 5O2 + 2Mn2+ + 8H2O + 2K+
From the equation, we can see that the mole ratio between H2O2 and KMnO4 is 5:2.
Since we have a 3.4g sample of the H2O2 solution, we need to find the mass of H2O2 in the solution. The mass of H2O2 can be calculated using the percentage composition:
Mass of H2O2 = (x/100) * mass of solution
= (x/100) * 3.4g
Now, we can convert the mass of H2O2 to moles:
Moles of H2O2 = Mass of H2O2 / Molar mass of H2O2
The molar mass of H2O2 (hydrogen peroxide) is 34.02 g/mol.
Now, using the mole ratio, we can calculate the moles of KMnO4 required:
Moles of KMnO4 = (5/2) * Moles of H2O2
Now, we have the moles of KMnO4 required and the volume of KMnO4 solution required (which is x mL). To find the molarity of the KMnO4 solution, we can use the formula:
Molarity (M) = Moles of solute / Volume of solution (in liters)
Since the volume of the KMnO4 solution is given in mL, we need to convert it to liters:
Volume of KMnO4 solution (in liters) = x mL / 1000
Now, we can calculate the molarity of the KMnO4 solution:
Molarity (M) = Moles of KMnO4 / Volume of KMnO4 solution (in liters)
To find the correct option, we need to substitute the values of x and solve for the molarity of the KMnO4 solution.
To find the molarity of the KMnO4 solution, we need to use the given information about the H2O2 solution.
First, let's understand the equation for the reaction between H2O2 and KMnO4:
2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5O2
In this equation, KMnO4 is the oxidizing agent, and it reacts with H2O2 (the reducing agent) to produce the products shown.
The balanced equation tells us that 2 moles of KMnO4 react with 3 moles of H2O2. Based on this, we can set up a proportion:
(2 moles of KMnO4) / (3 moles of H2O2) = (x milliliters of KMnO4 solution) / (3.4 grams of H2O2)
We are given that the mass of the H2O2 solution is 3.4 grams, so we'll use this value.
Now, let's rearrange the equation to solve for x, the volume of KMnO4 solution in milliliters:
x = (2 moles of KMnO4 * 3.4 grams of H2O2) / (3 moles of H2O2)
Next, we need to calculate the number of moles of H2O2 in the 3.4 grams sample. The molar mass of H2O2 is 34 g/mol, so we can use this value to convert from grams to moles:
moles of H2O2 = (3.4 grams of H2O2) / (34 g/mol)
Finally, substituting the values, we can calculate the volume of KMnO4 solution in milliliters:
x = (2 moles of KMnO4 * (3.4 grams of H2O2 / 34 g/mol)) / (3 moles of H2O2)
Simplifying the equation gives us the value of x. The value of x should be equal to the molarity of the KMnO4 solution.
Now, let's calculate the value of x:
x = (2 * (3.4 / 34)) / 3 = 0.2
Therefore, the molarity of the KMnO4 solution is 0.2 M.
So, the correct answer is (d) 0.2.
I think I know what the author of the problem expects; however, I don't think this is a viable problem BECAUSE you have two unknowns. That is you have x mL KMnO4 as well as y% H2O2 in the 3.4 g sample. That author assumes the x%(which is in grams) and x mL is the same and they can't be. But I think the following is what the author expects so here is the solution.
2KMnO4 + 5H2O2 + 3H2SO4 =>2MnSO4 + K2SO4 + 8H2O + 5O2
Assuming the x values CAN be the same let's try the 4 answers and see if they fit. For 1 M, we have 1M(xmL) = 1 millimils KMnO4. That will equal (5x/2) mmols H2O2 and (5x/2)*0.034(the millimolar mass H2O2) = 0.085x g H2O2. Then
[(0.085x g)/3.4]*100 = x%. Check it out and the left side doesn't equal the right side.
Do the same for 0.5M, 0.4M and 0.2M
One of them matches. Post your work if you get stuck.