1)If 36 mL of KMnO4 solution are required to oxidize 25 mL of 0.02932 M NaC204 solution, what is the concentration of the solution?
2)A Certain brand of iron supplement contains FeSO4-7H2O with miscellaneous binders and fillers. Suppose 22.93 mL of the KMn04 solution used in question one above are needed to oxidize Fe2+ to Fe3+ in a 0.4927 g pill. What is the mass % of FeSO4-7H2O(MM 278.03 g mol) in the pill?
1. Write the equation and balance it. I think 2 KMnO4 requires 5 Na2C24(note your typo).
moles KMnO4 = M x L = ??
Using the coefficients in the balanced equation, convert moles KMnO4 to moles Na2C2O4.
Now, M Na2C2O4 = moles Na2C2O4/L Na2C2O4.
2. Same thing for Fe sample. However, instead of M of Fe at the end, convert moles to grams FeSO4.7H2O, then
% = (mass FeSO4.7H2O/mass sample)*100 =xx
Post your work if you get stuck.
To find the concentration of the KMnO4 solution in question 1, we can use the concept of stoichiometry and molarity.
1) Here's how you can find the concentration of the KMnO4 solution:
Step 1: Calculate the number of moles of NaC204 in the given solution.
First, we need to find the moles of NaC204 using its molarity and volume.
moles = molarity * volume
moles of NaC204 = 0.02932 mol/L * 0.025 L = 0.000733 moles
Step 2: Use the stoichiometry of the reaction between KMnO4 and NaC204 to find the moles of KMnO4.
The balanced equation is:
2KMnO4 + 10NaC204 + 8H2SO4 -> 2MnSO4 + 5(Na)2SO4 + 10CO2 + 8H2O
According to the equation, 10 moles of NaC204 react with 2 moles of KMnO4.
So, moles of KMnO4 = (0.000733 moles NaC204) * (2 moles KMnO4 / 10 moles NaC204) = 0.0001466 moles
Step 3: Calculate the concentration of the KMnO4 solution.
Concentration (Molarity) = moles / volume
Concentration of KMnO4 = 0.0001466 moles / 0.036 L = 0.004072 M
Therefore, the concentration of the KMnO4 solution is 0.004072 M.
For question 2, you can find the mass percentage of FeSO4-7H2O in the pill using the given information and the concept of stoichiometry.
2) Here's how you can find the mass percentage of FeSO4-7H2O in the pill:
Step 1: Calculate the number of moles of KMnO4 used in the oxidation.
We are given the volume of KMnO4 used (22.93 mL), and we already calculated the concentration of KMnO4 (0.004072 M) in question 1.
Moles of KMnO4 = concentration * volume
Moles of KMnO4 = 0.004072 mol/L * 0.02293 L = 0.0000936 moles
Step 2: Use the stoichiometry of the reaction between KMnO4 and FeSO4-7H2O to find the moles of FeSO4-7H2O.
The balanced equation is:
5FeSO4 + MnO4^- + 8H+ -> 5Fe^3+ + Mn^2+ + 4H2O
According to the equation, 1 mole of KMnO4 reacts with 5 moles of FeSO4-7H2O.
So, moles of FeSO4-7H2O = (0.0000936 moles KMnO4) * (5 moles FeSO4-7H2O / 1 mole KMnO4) = 0.000468 moles
Step 3: Calculate the mass of FeSO4-7H2O in the pill.
Mass = moles * molar mass
Mass = 0.000468 moles * 278.03 g/mol = 0.1296 g
Step 4: Calculate the mass percentage of FeSO4-7H2O in the pill.
Mass percentage = (mass of FeSO4-7H2O / total mass of the pill) * 100%
Mass percentage = (0.1296 g / 0.4927 g) * 100% = 26.29%
Therefore, the mass percentage of FeSO4-7H2O in the pill is 26.29%.