A 10g sample of impure iron iii sulphate, FeSO4 was dissolved in water to make 250cm3 of solution. Exactly 25.0cm3 of the solution was titrated with 0.025mol/dm3 potassium manganate vii, KMnO4 until the solution just turned pink. The volume of potassium manganate vii that reacted with 25.0cm3 of the solution was 23.5cm3
The reaction equation is shown:
Mn0+H+Fe=Mn+Fe+H2O
a) Calculate the number of moles in 23.5cm3 of 0.025mol/dm3 potassium manganate vii
b) Calculate the number of moles of iron sulphate in 25.0cm3 of solution
c) Use your answer to (b) to calculate the number of moles of iron ii sulphate in 250cm3 of solution.
a) The number of moles in 23.5 cm3 of 0.025 mol/dm3 potassium manganate VII can be calculated using the formula:
moles = concentration x volume
moles = 0.025 mol/dm3 x 23.5 cm3
First, we need to convert cm3 to dm3:
23.5 cm3 = 23.5/1000 dm3 = 0.0235 dm3
moles = 0.025 mol/dm3 x 0.0235 dm3
moles = 0.0005875 moles
Therefore, there are 0.0005875 moles in 23.5 cm3 of the potassium manganate VII solution.
b) The number of moles of iron sulfate in 25.0 cm3 of the solution can be determined by the stoichiometry of the reaction:
1 mole of KMnO4 reacts with 1 mole of FeSO4
Since the moles of KMnO4 are known (0.0005875 moles), the number of moles of FeSO4 can be calculated as well:
moles of FeSO4 = moles of KMnO4 = 0.0005875 moles
Therefore, there are 0.0005875 moles of iron sulfate in 25.0 cm3 of the solution.
c) To calculate the number of moles of iron II sulfate in 250 cm3 of the solution, we can use the equation:
moles of FeSO4 = (moles of FeSO4 in 25.0 cm3) x (total volume of solution/ volume of solution used in titration)
moles of FeSO4 = (0.0005875 moles) x (250 cm3/ 25.0 cm3)
moles of FeSO4 = (0.0005875 moles) x (10)
moles of FeSO4 = 0.005875 moles
Therefore, there are 0.005875 moles of iron II sulfate in 250 cm3 of the solution.
a) To calculate the number of moles in 23.5 cm3 of 0.025 mol/dm3 potassium manganate vii, we can use the formula:
Moles = Concentration × Volume
Moles = 0.025 mol/dm3 × 23.5 cm3
However, we need to convert the volume from cm3 to dm3 before performing the calculation. Since 1 dm3 = 1000 cm3, we have:
Volume (dm3) = 23.5 cm3 / 1000 cm3/dm3 = 0.0235 dm3
Moles = 0.025 mol/dm3 × 0.0235 dm3 = 0.0005875 moles
Therefore, there are 0.0005875 moles of potassium manganate vii in 23.5 cm3 of the solution.
b) To calculate the number of moles of iron sulphate in 25.0 cm3 of the solution, we need to use the balanced equation:
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
From the equation, we see that the mole ratio between potassium manganate vii and iron sulphate is 1:5. This means that for every 1 mole of potassium manganate vii, there are 5 moles of iron sulphate.
Since there are 0.0005875 moles of potassium manganate vii in the titration, we can calculate the moles of iron sulphate:
Moles of iron sulphate = 0.0005875 moles × (5 moles of iron sulphate / 1 mole of potassium manganate vii)
Moles of iron sulphate = 0.0005875 moles × 5 = 0.0029375 moles
Therefore, there are 0.0029375 moles of iron sulphate in 25.0 cm3 of the solution.
c) To calculate the number of moles of iron ii sulphate in 250 cm3 of the solution, we can use the ratio between the volumes of the solutions used in the titration:
Moles of iron ii sulphate = Moles of iron sulphate × (Volume of solution in titration / Volume of the entire solution)
Moles of iron ii sulphate = 0.0029375 moles × (25.0 cm3 / 250 cm3)
Moles of iron ii sulphate = 0.0029375 moles × 0.1
Moles of iron ii sulphate = 0.00029375 moles
Therefore, there are 0.00029375 moles of iron ii sulphate in 250 cm3 of the solution.
To solve this problem, we need to use the given information and equations to calculate the number of moles of potassium manganate VII (KMnO4) and the number of moles of iron sulfate (FeSO4) in the solution.
a) Calculate the number of moles in 23.5 cm3 of 0.025 mol/dm3 potassium manganate VII:
The concentration of potassium manganate VII is given as 0.025 mol/dm3. This means that for every 1 dm3 of solution, there are 0.025 moles of potassium manganate VII.
To find the number of moles in 23.5 cm3, we need to convert cm3 to dm3:
1 dm3 = 1000 cm3
23.5 cm3 = 23.5/1000 dm3 = 0.0235 dm3
Now we can calculate the number of moles using the formula:
Number of moles = Concentration x Volume
Number of moles = 0.025 mol/dm3 x 0.0235 dm3 = 0.0005875 moles
So, there are 0.0005875 moles of potassium manganate VII in 23.5 cm3 of the solution.
b) Calculate the number of moles of iron sulfate (FeSO4) in 25.0 cm3 of solution:
We have the volume of the solution (25.0 cm3). To find the number of moles, we need to use the reaction equation and the volume of potassium manganate VII that reacted.
Based on the given equation:
MnO4+H+Fe=Mn+Fe+H2O
It shows that for every 1 mole of potassium manganate VII, 1 mole of iron sulfate reacts. So the number of moles of iron sulfate is the same as the number of moles of potassium manganate VII, which is 0.0005875 moles.
Therefore, there are 0.0005875 moles of iron sulfate in 25.0 cm3 of solution.
c) Use the answer to part (b) to calculate the number of moles of iron II sulfate in 250 cm3 of solution:
We have the number of moles of iron sulfate (FeSO4) in 25.0 cm3 of solution, which is 0.0005875 moles. To find the number of moles in 250 cm3 of solution, we can use the following proportion:
moles in 25.0 cm3 / volume of 25.0 cm3 = moles in 250 cm3 / volume of 250 cm3
0.0005875 moles / 25.0 cm3 = moles / 250 cm3
Cross multiplying:
moles = (0.0005875 moles / 25.0 cm3) x 250 cm3
Calculating:
moles = 0.0005875 moles x 250 cm3 / 25.0 cm3
moles = 0.005875 moles
Therefore, there are 0.005875 moles of iron II sulfate in 250 cm3 of solution.