calculate the volume of water that can be treated if 50 g of potassium permanganate is added to the system. Assume that the concentrated iron (II) ions in well water is 5.0ppm
To calculate the volume of water that can be treated, we need to consider the reaction between potassium permanganate and iron (II) ions. The balanced chemical equation for this reaction is:
5 Fe2+ + 2 MnO4- + 6 H+ -> 5 Fe3+ + 2 Mn2+ + 3 H2O
From the equation, we can see that for every 2 moles of potassium permanganate (MnO4-) used, 2 moles of iron (II) ions (Fe2+) are consumed.
Firstly, we need to convert the concentration of iron (II) ions from parts per million (ppm) to moles per liter (mol/L). To do this, we use the molar mass of iron (II) ions.
The molar mass of Fe2+ = 55.845 g/mol
Converting 5 ppm of iron (II) ions to molarity:
Mass of Fe2+ = Concentration (ppm) * Volume * Molar mass / 10^6
Where:
- Mass of Fe2+ = 5 ppm * unknown volume * 55.845 g/mol / 10^6 g
Once we determine the mass of iron (II) ions, we can calculate the moles of Fe2+:
Moles of Fe2+ = Mass of Fe2+ / Molar mass of Fe2+
Now, since the ratio between Fe2+ and MnO4- is 2:2, we can infer that the moles of MnO4- also need to be the same as Fe2+.
Moles of MnO4- = Moles of Fe2+
Given that 50 g of potassium permanganate (KMnO4) is added to the system, we can calculate the moles of MnO4-:
Moles of MnO4- = Mass of KMnO4 / Molar mass of KMnO4
Molar mass of KMnO4 = 39.098 g/mol (potassium) + 54.938 g/mol (manganese) + 15.999 g/mol (oxygen, 4 atoms)
Now, with the moles of MnO4- and Fe2+ obtained, we can determine the volume of water treated.
Volume of water treated = Moles of MnO4- * Molar volume of water
The molar volume of water at standard temperature and pressure (STP) is approximately 22.4 L/mol.
Therefore, by following these calculations, we can determine the volume of water that can be treated using 50 g of potassium permanganate.