9.8g of iron(II)salt FeSO4.(NH4)2.6H2O was dissolved in dilute sulphuric(VI)acid.Distilled water was added to it and made up to 250cm³.25cm³ of this solution was pippeted and transferred to a conical flask.It was found that this solution required 24.0cm³ of potassium manganate (VII) solution for complete neutralization.
MnO4+8H+5Fe~Mn+5Fe+4H2O
a)i) Calculate the molarity of iron(II) salt
ii) Calculate the moles of Fe in 25.0cm³ solution
iv)Calculate the molarity of the MnO4
i) To calculate the molarity of iron(II) salt, we need to use the equation:
molarity (M) = moles/volume (in liters)
First, we need to determine the moles of iron(II) salt. The molar mass of FeSO4.(NH4)2.6H2O can be calculated as follows:
FeSO4: 1 Fe x 55.845 g/mol = 55.845 g/mol
(NH4)2.6H2O:
2 NH4 x 18.04 g/mol = 36.08 g/mol
6 H2O x 18.02 g/mol = 108.12 g/mol
Total molar mass = 55.845 g/mol + 36.08 g/mol + 108.12 g/mol = 200.045 g/mol
Now, we can calculate the moles of iron(II) salt:
moles = mass/molar mass
moles = 9.8 g / 200.045 g/mol = 0.049 moles
Next, we need to convert the volume from cm³ to liters:
volume = 25.0 cm³ * (1 L/1000 cm³) = 0.025 L
Finally, we can calculate the molarity of iron(II) salt:
molarity = moles/volume
molarity = 0.049 moles/0.025 L
molarity ≈ 1.96 M
ii) We already know that the volume is 25.0 cm³. To calculate the moles of Fe in 25.0 cm³ of solution, we can use the molarity of iron(II) salt we just calculated:
moles = molarity * volume
moles = 1.96 M * 0.025 L
moles = 0.049 moles
iii) In the balanced equation, we see that the ratio between Fe and MnO4 is 1:1. Therefore, the moles of Fe is equal to the moles of MnO4. We just calculated that the moles of Fe in 25.0 cm³ of solution is 0.049 moles, so the molarity of MnO4 is also 0.049 M.