2 KMnO4 + 5 H2C2O4 + 3 H2SO4 ---> 2 MnSO4 + 10 CO2 + 8 H2O + K2SO4
How many milliliters of a 0.123 M KMnO4 solution is needed to react completely with 3.141 of oxalic acid?
Please help I've worked this every way I know.
This four step process will just about work them all. Print this and save it.
1. Write and balance the equation. You've done that.
2. mols H2C2O4 = grams/molar mass = ?
3. Using the coefficients in the balanced equation, convert mols H2C2O4 5o mols KMnO4. That will be mols H2C2O4 x 2/5 = ?
4. M KMnO4 = mols KMnO4/L KMnO4
You know mols and M, solve for L and convert to mL.
Post your work if you get stuck but remember this 4-step process.
Ok so you take 3.141g and divide that by 90.036?
Yes, for step 2.
Thanks. The problem I was having, I would multiply the grams by the molar mass by everything else I done was correct to what you were saying. What I don't understand is why you divide and not multiple(I do this for when it says a volume of somthing and get the right answer)
I don't understand much of what you have written. For the mols oxalic acid, it is grams = mols x molar mass. If you know molar mass and grams and you want mols, that is
g = mols x molar mass
g/molar mass = mols x molar mass/molar mass
Now on the right side, molar mass in the numerator cancels with the molar mass in the denominator to leave just mols and that makes mols = grams/molar mass.
If you give an example of what you are talking about with the volume, perhaps I can help with that too. I think your problem is one of algebra.
If g = mols x molar mass, then
g/molar mass must = mols.
It's 0.854 M NaOH solution react completely with 2.1 L of a 0.972 M BaCl2. It's a 2:1 but I see what I did. Late nights sometimes mess with ya.
Thank you. Might need help In the future.
glad to help