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 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.
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