Achemistry student weighs out 0.0975 g of acrylic acid (hch2chco2) into a 250. ml volumetric flask and diluted to the mark with distilled water. he plans to titrate the acid with 0.0500 m naoh solution. calculate the volume of naoh solution the student will need to add to reach the equilvalence point. be sure your answer has the correct number of significant digits.

27.06 mL.

Explanation:

Molarity is the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of acrylic acid)/(V of the solution (L))

M = (mass/molar mass)acrylic acid / (V of the solution (L))

mass of acrylic acid = 0.0975 g, molar mass of acrylic acid = 72.06 g/mol, V of the solution = 250 mL = 0.25 L.

∴ M = (0.0975 g/72.06 g/mol)/(0.25 L) = 0.0054 M.

The no. of millimoles of acid is equal to that of the base at the neutralization.

∴ (XMV) NaOH = (XMV) acrylic acid.

X is the no. of reproducible H⁺ (for acid) or OH⁻ (for base),

M is the molarity.

V is the volume.

X = 1, M = 0.05 M, V = ??? mL.

X = 1, M = 0.0054 M, V = 250.0 mL.

∴ V of NaOH = (XMV) acrylic acid/(XM) NaOH = (1)(0.0054 M)(250.0 mL)/(1)(0.05 M) = 27.06 mL.

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