A 0.2mol/L solution of KMnO4 has been added to 10 ml of 0.5mol/L acidified FeSO4.
write the balance equation of the reacttion then calculate the volume of the oxidizing solution that should be added to oxidize all the Fe^2+ ions.
please help and thank you!!
FeSO4 + KMnO4 + H^+ ==> Fe2(SO4)3 + K2SO4 + MnSO4 + H2O.
Fe is oxidized from +2(each atom) to +3 while Mn is reduced from +7 to +2. I assume you can balance from here.
Use the coefficients in the balanced equation to convert mole Fe to moles KMnO4. Post your work if you get stuck.
Sure, I can help you with that! But fair warning, my answers may contain a (a) grin 😄. Now, let's get down to business!
The balanced equation for the reaction between KMnO4 and FeSO4 can be written as:
2 KMnO4 + 10 FeSO4 + 8 H2SO4 → K2SO4 + 2 MnSO4 + 8 H2O + 5 Fe2(SO4)3
Now, to determine the volume of the KMnO4 solution needed to oxidize all the Fe^2+ ions, we'll use stoichiometry! Pay attention, there won't be any jokes here, just plain chemistry math 😉
First, let's find the number of moles of FeSO4 in the 10 mL solution:
0.5 mol/L × 0.01 L = 0.005 moles of FeSO4
According to the balanced equation, 10 moles of FeSO4 react with 2 moles of KMnO4. Therefore, the number of moles of KMnO4 required would be:
0.005 moles of FeSO4 × (2 moles of KMnO4 / 10 moles of FeSO4) = 0.001 moles of KMnO4
Now we need to find the volume of the 0.2 mol/L KMnO4 solution that contains 0.001 moles:
Volume (L) = Moles / Concentration
Volume (L) = 0.001 moles / 0.2 mol/L = 0.005 L or 5 mL
So, to oxidize all the Fe^2+ ions, you would need to add 5 mL of the 0.2 mol/L KMnO4 solution.
I hope that helps! If you have any more questions, feel free to ask. 🤡
The balanced equation for the reaction between KMnO4 and FeSO4 in acidic solution is:
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
To calculate the volume of the oxidizing solution (KMnO4) required to react with all the Fe2+ ions, we need to use the stoichiometry of the balanced equation.
From the equation, we can see that the ratio of KMnO4 to Fe2+ is 1:5. This means that for every 1 mole of KMnO4, it reacts with 5 moles of Fe2+ ions.
Given that we have a 0.2 mol/L solution of KMnO4 and 10 mL of a 0.5 mol/L solution of FeSO4, we can use the following steps to calculate the volume of KMnO4 solution required:
1. Calculate the number of moles of Fe2+ in the initial solution:
Moles of Fe2+ = (0.5 mol/L) x (0.01 L) = 0.005 mol
2. Since the molar ratio of KMnO4 to Fe2+ is 1:5, we can calculate the number of moles of KMnO4 required to react with all the Fe2+ ions:
Moles of KMnO4 = 5 x Moles of Fe2+ = 5 x 0.005 mol = 0.025 mol
3. Calculate the volume of the KMnO4 solution required:
Volume of KMnO4 = Moles of KMnO4 / Concentration of KMnO4
Volume of KMnO4 = 0.025 mol / 0.2 mol/L = 0.125 L or 125 mL
Therefore, the volume of the oxidizing solution (KMnO4) that should be added to oxidize all the Fe2+ ions is 125 mL.
To write the balanced equation for the reaction between KMnO4 and FeSO4, we first need to determine the oxidation states of the elements involved:
- In KMnO4, the oxidation state of potassium (K) is +1, and the oxidation state of oxygen (O) is -2. Since there are four oxygen atoms, the total oxidation state contributed by oxygen is -8. Therefore, the oxidation state of manganese (Mn) must be +7 to balance the total charge of -8.
- In FeSO4, the oxidation state of sulfur (S) is +6 (from the -2 charge in SO4), and the overall charge of FeSO4 must be zero. Thus, the oxidation state of iron (Fe) is +2.
Based on these oxidation states, the balanced equation for the reaction can be written as follows:
2 KMnO4 + 10 FeSO4 + 8 H2SO4 -> K2SO4 + 2 MnSO4 + 5 Fe2(SO4)3 + 8 H2O
Now, to calculate the volume of the oxidizing solution (KMnO4) required to oxidize all the Fe2+ ions in the 10 mL of the 0.5 M FeSO4 solution, we need to consider the stoichiometry of the reaction. From the balanced equation, we know that 2 moles of KMnO4 react with 10 moles of FeSO4.
Given that the initial FeSO4 solution has a concentration of 0.5 mol/L and a volume of 0.010 L (10 mL), we can calculate the number of moles of FeSO4:
moles of FeSO4 = concentration * volume = 0.5 * 0.010 = 0.005 mol
According to the stoichiometry of the reaction, 10 moles of FeSO4 react with 2 moles of KMnO4. Therefore, the number of moles of KMnO4 required can be calculated as:
moles of KMnO4 = (moles of FeSO4 / 10) * 2 = (0.005 / 10) * 2 = 0.001 mol
Finally, we can determine the volume of the KMnO4 solution by using its concentration:
volume of KMnO4 solution = moles of KMnO4 / concentration = 0.001 / 0.2 = 0.005 L
So, the volume of the oxidizing solution needed to oxidize all the Fe2+ ions in 10 mL of 0.5 M acidified FeSO4 is 0.005 L (or 5 mL).