What volume of O.1500 KMnO4 would be required to react with 1 g of formic acid HCO2H, in acid solution ?

Question

The unbalanced equation for the problem is
MnO4(-)+HCO2H-->Mn(2+)+Co2
Ok, i know that the first couple of steps involve balancing the reaction in acidic conditions, but how does that help? In addition to that , what else do you need to do ?

Answer 1

First, balance the equation. Permanginate ions form manganese dioxide and the organic compound is going to be oxidized to CO2 and H2O:

MnO4 + 2H2CO2 -> MnO2 + 2H2O + 2CO2

Now, notice the relationships between the numbers of molecules, especially the permanginate and formic acid. One molecule of permanginate is going to oxidize two molecules of formic acid.

Next, calculate the molarity. What applies to molecules applies to moles, too. "Moles" are actually numbers. 1 mole of permanginate will oxidize two moles of formic acid.

Therefore, the next thing to calculate is how many moles of formic acid 1 gram represents. Besides being a number, a mole also is a particular weight for a particular compound. 1 mole of formic acid weighs what 1 mole of each of its individual atoms weighs. A periodic table is the place to find atomic weights. They are usually the numbers with decimal points. Formic acid is two oxygens, a pair of hydrogens and a carbon atom. This works out to be 2*16 + 2*1 + 12 = 46 grams per mole. (I rounded the answers). Therefore, 1 gram of formic acid is 1/46 mole, or 0.022 moles.

Now, 0.022 moles of formic acid represents twice the number of moles of permanginate needed to oxidize it. This is of course 0.011 moles of permanginate. The next thing to do is figure out what this actually weighs. Permanginate actually exists as a potassium salt, KMnO4 and the molecular weight is 158 grams per mole, meaning of course 0.011 moles is 1.738 grams. Finally, one has to determine the volume of a 0.1500 molar solution of KMnO4 which contains 1.738 grams of KMnO4. 0.1500 moles of permanginate weigh 23.7 grams. This is how many grams of permanginate are in 1 liter of this solution (1000 mls). How many mls of the 0.1500 M permanginate contain 1.738 grams? This is the answer!

Answer 2

Mr(Formic Acid)
C x 1 = 12 x 1 = 12
H x 2 = 1 x 2 = 2
O x 2 = 16 x 2 = 32
12 + 2 + 32 = 46
moles (Methanoic Acid) = 1/46 = 0.0217 mol.
As methanoic acid is 'an acid' it provides the acid solution in its own right.
KMnO4 + HCO2H = HMnO4 + K+HCO2-
Molar Ratios are 1:1::1:1
So moles methanoic acid = moles potassium permangate at 0.0217.
vol = moles x 1000/conc
vol = 0.0217 x 1000/0.15 = 144.666..cm^3

NB Formic Acid is an archaic name. under the modern IUPAC nomenclature it is now METHANOIC ACID.