Asked by Anna
A solution of hydrogen peroxide, H2O2, is titrated with a solution of potassium permanganate, KMnO4.
It requires 51.7 mL of 0.145 M KMnO4 to titrate 20.0 g of the solution of hydrogen peroxide. What is the mass percentage of H2O2 in the solution?
If you want the whole equation I'll type it but it's long. I think all you need is the mole ratio which is 5 mol H202 : 2 mol KMnO4.
so far I changed 51.7 mL to L. Got .00750 mol KMnO4 using molarity and volume. converted that to .0188 mol H2O2.
It requires 51.7 mL of 0.145 M KMnO4 to titrate 20.0 g of the solution of hydrogen peroxide. What is the mass percentage of H2O2 in the solution?
If you want the whole equation I'll type it but it's long. I think all you need is the mole ratio which is 5 mol H202 : 2 mol KMnO4.
so far I changed 51.7 mL to L. Got .00750 mol KMnO4 using molarity and volume. converted that to .0188 mol H2O2.
Answers
Answered by
DrBob222
You may have almost finished it then but I'll start from the beginning.
Convert M x L KMnO4 to mols.
Convert mols KMnO4 to mols H2O2 using the coefficients in the balanced equation.
Convert mols H2O2 to grams H2O2.
Then %H2O2 = [grams H2O2/mass of sample]*100.
Mass of the sample is 20.0 g in the problem.
Convert M x L KMnO4 to mols.
Convert mols KMnO4 to mols H2O2 using the coefficients in the balanced equation.
Convert mols H2O2 to grams H2O2.
Then %H2O2 = [grams H2O2/mass of sample]*100.
Mass of the sample is 20.0 g in the problem.
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