If 36.00 mL of KMnO4 are required to oxidize 25.00 mL of 0.02932 M NaC2O4 solution, what is the concentration of KMnO4?
I believe this has been answered above.
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To find the concentration of KMnO4, we need to use the given information and apply the concept of stoichiometry.
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions.
First, let's determine the stoichiometry of the reaction between KMnO4 and NaC2O4. The balanced equation for the reaction is:
2KMnO4 + 5NaC2O4 + 8H2SO4 → 2MnSO4 + 10CO2 + K2SO4 + 8H2O + 5Na2SO4
From the balanced equation, we can see that it takes 2 moles of KMnO4 to react with 5 moles of NaC2O4. This can be simplified to the ratio of 2:5.
Now, we can use the given information to set up a stoichiometry calculation:
Given:
Volume of KMnO4 solution (V1) = 36.00 mL
Concentration of NaC2O4 solution (C2) = 0.02932 M
Volume of NaC2O4 solution (V2) = 25.00 mL
We need to convert the volume of KMnO4 solution to moles:
Moles of KMnO4 (n1) = V1 * C1
where C1 is the unknown concentration of KMnO4.
Using the stoichiometric ratio of 2:5, we can set up the equation:
n1/n2 = 2/5
Plugging in the values:
(V1 * C1) / (V2 * C2) = 2/5
Rearranging the equation to solve for C1:
C1 = (2/5) * (V2 * C2) / V1
Now we can substitute the given values to find the concentration of KMnO4:
C1 = (2/5) * (25.00 mL * 0.02932 M) / 36.00 mL
Calculating this expression will give you the concentration of KMnO4 in Molarity (M).