36.85 mL of a KMnO4 solution when acidified reacted with 1.076g of Mohr's salt, Fe(NH4)2(SO4)2 x 6H2O What is the normality of the KMnO4 solution?
How do I set this up? I can't figure it out at all...
MnO4- + 5Fe2+ ==> Fe3+ + Mn2+
N = #milliequivalents/mL = #me/mL.
#me = grams/mew (where mew = milliequivalent weight)
#me = g/mew = 1.076/0.39214 = 2.7439
N = me/mL = 2.7439/36.85 = ??
Note: mew of Mohr's salt = molar mass/delta e. 5 moles Fe undergo 5e change which is 1e/1 mol so molar mass = equivalent mass (equivalent weight).
no solution
To calculate the normality of the KMnO4 solution, you need to follow a series of steps. Here's how you can set it up:
Step 1: Determine the balanced chemical equation for the reaction between KMnO4 and Mohr's salt. The equation is as follows:
5Fe(NH4)2(SO4)2 ⋅ 6H2O + KMnO4 + 8H2SO4 → 5Fe2(SO4)3 + MnSO4 + K2SO4 + 8H2O + 10NH4HSO4
Step 2: Use the given mass of Mohr's salt (1.076g) to calculate the number of moles of Mohr's salt. The molar mass of Mohr's salt is 392.14 g/mol.
moles of Mohr's salt = mass of Mohr's salt / molar mass of Mohr's salt
Step 3: From the balanced chemical equation, determine the stoichiometric ratio between Mohr's salt and KMnO4. In this case, it is 1:1, as both reactants have a coefficient of 1.
Step 4: Since the stoichiometry is 1:1 and the number of moles of Mohr's salt is equivalent to the number of moles of KMnO4, we can conclude that the number of moles of KMnO4 is also equal to the number of moles of Mohr's salt.
moles of KMnO4 = moles of Mohr's salt
Step 5: Calculate the volume in liters of the KMnO4 solution using the provided volume (36.85 mL).
volume of KMnO4 solution (in liters) = volume of KMnO4 solution (in mL) / 1000
Step 6: Calculate the normality of the KMnO4 solution using the formula:
Normality (N) = moles of KMnO4 / volume of KMnO4 solution (in liters)
That's it! You have set up the problem and can now proceed to perform the calculations step-by-step.
To determine the normality of the KMnO4 solution, you need to set up a balanced chemical equation for the given reaction. From the information provided, we know that KMnO4 solution reacts with Mohr's salt (Fe(NH4)2(SO4)2 x 6H2O) in an acidified medium.
The balanced chemical equation for this reaction is as follows:
5 Fe(NH4)2(SO4)2 x 6H2O + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + 5(NH4)2SO4 + K2SO4 + 2 MnSO4 + 8 H2O
Now, let's calculate the number of moles of Mohr's salt used in the reaction:
1.076g of Mohr's salt / molar mass of Mohr's salt
- The molar mass of Mohr's salt (Fe(NH4)2(SO4)2 x 6H2O) can be calculated as follows:
Molar mass of Fe = 55.85 g/mol
Molar mass of N = 14.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of water (H2O) = 18.02 g/mol
Molar mass of Mohr's salt = (2*55.85) + (2*14.01) + (2*32.07) + (12*1.01) + (6*18.02) = 392.26 g/mol
Therefore, the number of moles of Mohr's salt used in the reaction is:
1.076g / 392.26 g/mol
Next, we need to determine the mole ratio between KMnO4 and Mohr's salt, which is obtained from the balanced equation. In this case, it is 2:1, which means that 2 moles of KMnO4 react with 1 mole of Mohr's salt.
Now, let's calculate the number of moles of KMnO4 used in the reaction:
(Number of moles of Mohr's salt * 2) / 1
Next, we need to find the volume (in liters) of the KMnO4 solution used in the reaction. You mentioned that the volume is 36.85 mL, so we need to convert it to liters:
36.85 mL / 1000 mL/L
Now you have the number of moles of KMnO4 and the volume of the solution used. To find the normality, you need to divide the number of moles of KMnO4 by the volume of the solution in liters.
Normality = (Number of moles of KMnO4) / (Volume of KMnO4 solution in liters)
Simply plug in the values obtained from the calculations, and you will have the normality of the KMnO4 solution.