A 25cm aliquot of ammonium iron 2 tetraoxomanganate 7 containing 39.2g of its salt in 250cm solution, needed 25cm of potassium tetraoxomanganate 7 solution for a colour change to occur. What is the concentration of the potassium tetraoxomanganate 7 solution?
Do you have a formla for this material; i.e., ammonium iron 2 tetraoxomanganate 7? Do you have an equation for the readction with KMnO4.
i need the answer
To find the concentration of the potassium tetraoxomanganate (VII) solution, we need to use the given information.
Step 1: Calculate the moles of ammonium iron(II) tetraoxomanganate(VII):
Given mass = 39.2 g
Molar mass of ammonium iron(II) tetraoxomanganate(VII) = 392.02 g/mol
Moles = mass / molar mass
Moles = 39.2 g / 392.02 g/mol
Moles = 0.1 mol
Step 2: Calculate the concentration of ammonium iron(II) tetraoxomanganate(VII) in the solution:
Given volume = 250 cm³
Concentration (in mol/dm³) = moles / volume (in dm³)
Volume (in dm³) = 250 cm³ / 1000 cm³/dm³
Volume (in dm³) = 0.25 dm³
Concentration = 0.1 mol / 0.25 dm³
Concentration = 0.4 mol/dm³
Step 3: Use the stoichiometry of the balanced equation to find the concentration of potassium tetraoxomanganate (VII):
From the balanced equation, we know that the molar ratio between ammonium iron(II) tetraoxomanganate(VII) and potassium tetraoxomanganate(VII) is 1:1.
So, the concentration of potassium tetraoxomanganate(VII) solution = concentration of ammonium iron(II) tetraoxomanganate(VII) solution.
Therefore, the concentration of the potassium tetraoxomanganate(VII) solution is 0.4 mol/dm³.
To find the concentration of the potassium tetraoxomanganate VII (KMnO4) solution, we will use the information provided in the question. Let's break down the problem step by step.
1. Calculate the amount of ammonium iron(II) tetraoxomanganate VII (NH4)2Fe(MnO4)7 in the 25 cm³ aliquot:
Given that the 250 cm³ solution contains 39.2 g of (NH4)2Fe(MnO4)7:
Therefore, the concentration (C1) of (NH4)2Fe(MnO4)7 can be found as:
C1 = (mass of (NH4)2Fe(MnO4)7) / (volume of solution)
C1 = 39.2 g / 250 cm³
2. Calculate the number of moles of (NH4)2Fe(MnO4)7:
To find the number of moles, we use the formula:
Moles = mass / molar mass
The molar mass of (NH4)2Fe(MnO4)7 can be calculated by summing up the molar masses of each atom it contains:
Molar mass of (NH4)2Fe(MnO4)7 = (2 × molar mass of NH4) + (molar mass of Fe) + (7 × molar mass of MnO4)
Now, calculate the moles:
Moles of (NH4)2Fe(MnO4)7 = mass / molar mass
3. Calculate the moles of KMnO4 used:
The balanced chemical equation for the reaction between (NH4)2Fe(MnO4)7 and KMnO4 is:
(NH4)2Fe(MnO4)7 + KMnO4 -> FeO4Mn + K2O4Mn + 2NH4NO3
The stoichiometry of the reaction indicates that 1 mole of (NH4)2Fe(MnO4)7 reacts with 1 mole of KMnO4.
Therefore, the moles of KMnO4 used in the reaction are equal to the moles of (NH4)2Fe(MnO4)7.
4. Calculate the concentration of KMnO4 solution:
You now have the number of moles of KMnO4 used and the volume of KMnO4 solution used (which is 25 cm³). You can calculate the concentration (C2) of KMnO4 using the formula:
C2 = moles of KMnO4 / volume of KMnO4 solution
C2 = moles of (NH4)2Fe(MnO4)7 / 25 cm³
Now, you have found the concentration of the KMnO4 solution. Plug in the values you calculated, and that will give you the answer.