A potassium permanganate solution containing 1.58 g of KMnO4 per liter is added into 5 ml of an acidified iron (||) sulfate solution .Decolorizing of KMnO4 solution ceases when 10ml of the solution has been added.
Calculate the mass of FeSo4.7H2O which has been dissolved in one liter of solution.
mols KMnO4 = grams KMnO4/molar mass KMnO4.
M KMnO4 = mols/L
Here is the redox part of the reaction and I've balanced that part. You can finish it but the redox part is all that matters.
5Fe^2+ + MnO4^- --> 5Fe^3+ + Mn^2+
mols KMnO4 used = M x L = ?
Using the coefficients in the balanced equation, convert mols KMnO4 to mols Fe^2+.
Then g FeSO4.7H2O = mols FeSO4.7H2O x molar mass FeSO4.7H2O and that's the grams in 5 mL of the solution. Convert that to g in 1 L.
Thx
To calculate the mass of FeSO4.7H2O dissolved in one liter of solution, we can use the stoichiometry of the reaction between potassium permanganate (KMnO4) and iron(II) sulfate (FeSO4).
The balanced chemical equation for the reaction is:
5 FeSO4 + 2 KMnO4 + 8 H2SO4 → 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O
According to the given information, the decolorizing of the KMnO4 solution ceases when 10 ml of the solution has been added. It means that the reaction has reached the stoichiometric point where all the FeSO4 has reacted with KMnO4.
To find the moles of KMnO4 added, we can use its molar mass:
Molar mass of KMnO4 = 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol) = 158.04 g/mol
The concentration of the KMnO4 solution is given as 1.58 g/L, which is equivalent to 1.58 g/1000 ml. Thus, the moles of KMnO4 added is:
moles of KMnO4 = (1.58 g/1000 ml) × (10 ml) / (158.04 g/mol) = 0.00998 mol
From the balanced equation, the stoichiometric ratio between KMnO4 and FeSO4 is 2:5. Therefore, the moles of FeSO4 reacted will be:
moles of FeSO4 = (0.00998 mol) × (5/2) = 0.02495 mol
The molar mass of FeSO4.7H2O is:
Molar mass of FeSO4.7H2O = (55.85 g/mol + 32.06 g/mol + 4(16.00 g/mol)) + 7(18.02 g/mol) = 278.01 g/mol
Finally, the mass of FeSO4.7H2O dissolved in one liter of solution is:
mass of FeSO4.7H2O = (0.02495 mol) × (278.01 g/mol) = 6.93 g
Therefore, the mass of FeSO4.7H2O which has been dissolved in one liter of the solution is approximately 6.93 grams.
To calculate the mass of FeSO4·7H2O (iron (II) sulfate heptahydrate) dissolved in one liter of solution, we can use the stoichiometry of the reaction between KMnO4 (potassium permanganate) and FeSO4 (iron (II) sulfate).
Here's the step-by-step process to find the mass of FeSO4·7H2O:
Step 1: Determine the moles of KMnO4 used
First, calculate the number of moles of KMnO4 used in the reaction by using the molarity of the KMnO4 solution and the volume used:
Molarity of KMnO4 solution = 1.58 g/L
Volume of KMnO4 used = 10 mL = 0.01 L
Moles of KMnO4 = Molarity × Volume
Moles of KMnO4 = 1.58 g/L × 0.01 L
Step 2: Determine the moles of FeSO4
The balanced equation for the reaction between KMnO4 and FeSO4 is:
2 KMnO4 + 10 FeSO4 + 8 H2SO4 → 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O
From the equation, we can see that the ratio of KMnO4 to FeSO4 is 2:10. Therefore,
Moles of FeSO4 = (Moles of KMnO4 × 10) / 2
Step 3: Calculate the mass of FeSO4·7H2O in one liter of solution
The molar mass of FeSO4·7H2O is 278.02 g/mol.
Mass of FeSO4·7H2O in one liter of solution = Moles of FeSO4 × Molar mass of FeSO4·7H2O
Mass of FeSO4·7H2O in one liter of solution = Moles of FeSO4 × 278.02 g/mol
So, by substituting the values we obtained from the previous steps, we can calculate the mass of FeSO4·7H2O in one liter of solution.