An analytical chemist weighs out of an unknown monoprotic acid into a volumetric flask and dilutes to the mark with distilled water. She then titrates this solution with solution. When the titration reaches the equivalence point, the chemist finds she has added of solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits.

The given question is incomplete. The complete question is as follows.

An analytical chemist weighs out 0.093 g of an unknown monoprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.1600 M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 6.5 mL of NaOH solution. Calculate the molar mass of the unknown acid. Round your answer to 2 significant digits.

Explanation:

The given data is as follows.

   Mass of monoprotic acid = 0.093 g,

   Volume of monoprotic acid = 250 mL,

  [NaOH] = 0.1600 M, and volume of NaOH required at reaches the equivalence point = 6.5 mL.

Now, first of all we will calculate the moles of NaOH as follows.

        Moles of NaOH = molarity × volume (L)

                             = 0.1600 M × 0.0065 L

                             = 0.00104 moles

Let us assume that monoprotic acid is HA. And, it is known that monoprotic acid has 1 hydrogen ion and NaOH is also monoprotic base, means it has 1 hydroxide ion. Therefore, the reaction will be as follows.

Hence, 1 mole of NaOH = 1 mole of HA

So, 0.00104 moles of NaOH = 0.00104 moles of HA

Now, we will calculate the molar mass of unknown acid as follows.

      Molar mass of unknown acid =

                                  =

                                  = 89.42 g/mol

Thus, we can conclude that molar mass of the unknown acid is 89.42 g/mol.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g