how would your calculation of the molarity of KMnO4 be affected if you had added twice as much sulfuric acid and water as was called for in the procedure?
4 KMnO4(aq) + 6 H2SO4(aq) = 2 K2SO4(aq) + 4 MnSO4(aq) + 6 H2O(l) + 5O2(g)
That would depend on what your limiting reagent is. If it was KMnO4, then the calculation should not change. If it was H2SO4, the calculation will change according to volume etc
However, since adding water will change the volume, the calculations should change as well
To calculate the molarity of KMnO4, the key parameters needed are the moles of KMnO4 and the volume of the solution. Adding extra sulfuric acid and water will not directly affect the molarity of KMnO4. However, it might impact the volume of the solution, which in turn could affect the molarity calculation.
Here's how you would calculate the molarity of KMnO4:
1. Determine the moles of KMnO4: This can be obtained using the formula:
Moles = Mass of KMnO4 (g) / Molar mass of KMnO4 (g/mol)
The molar mass of KMnO4 is calculated by adding up the atomic masses of potassium (K), manganese (Mn), and oxygen (O) based on their respective atomic weights from the periodic table.
2. Calculate the volume of the solution: This can be determined using the following equation:
Volume (L) = Volume of KMnO4 solution (mL) / 1000
If you have added extra sulfuric acid and water, it could lead to an increase in the final volume of the solution. Make sure to take this into account.
3. Compute the molarity: Molarity (M) is defined as moles of solute per liter of solution. Therefore, you can calculate it by dividing the moles of KMnO4 by the volume of the solution:
Molarity (M) = Moles of KMnO4 / Volume of Solution (L)
By following these steps, you can calculate the molarity of KMnO4 accurately, irrespective of the excess sulfuric acid and water added.