Question

Here is the information: A 1.2516 gram sample of a mixture of CaCO3 and Na2SO4 was analyzed by dissolving the sample and adding C2O42- to completely precipitate the Ca2+ as CaC2O4. The CaC2O4 was dissolved in sulfuric acid and the resulting H2C204 was titrated with a standard KMnO4 solution.
This is the balanced equation: 3H2C2O4 + 2H+ + MnO4- ---> Mn2+ + 4H20 + 6CO2
Another part: The titration of the H2C2H4 obtained required 35.62 milliliters of .1092 molar MnO4- solution. Calculate the number of moles of H2C2O4 that reacted with the MnO4-. i think the correct answer is .01167 mol H2C2O4. This is not my question.
Next Part: Calculate the number of moles of CaCO3 in the original sample. Based on the answer to the part above, the answer is .003883 mol CaCO3 (i think this is correct, however this is not my question)
Calculate the percentage by weight of CaCO3 in the original sample.
FInd this based off of the .01167 mol H2C2O4 and the .003883 mol CaCO3. Basically just use the answers above to solve for this.

Answer 1

Answer:

93,32 % (w/w)

Explanation:

The Ca²⁺ of CaCO₃ was completely precipitate to CaC₂O₄ that results in H₂C₂O₄. The titration of this one is:

3H₂C₂O₄ + 2H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O + 6CO₂

As you required 35,62mL of 0,1092M MnO₄⁻, the moles of H₂C₂O₄ are:

0,03562L×$\frac{0,1092M}{L}$ =

3,890x10⁻³mol of MnO₄⁻×$\frac{3molH_{2}C_{2}O_{4}}{1mol MnO_{4}^-}$ = 0,01167 mol of H₂C₂O₄

The number of moles of CaCO₃ are the same number of moles of H₂C₂O₄ because every Ca²⁺ was converted in CaC₂O₄ that was converted in H₂C₂O₄. That means: 0,01167 mol of CaCO₃

0,01167 mol of CaCO₃ are:

0,01167 mol of CaCO₃×$\frac{100,0869g}{1mol}$ = 1,168 g of CaCO₃

As the mass of the initial mixture is 1,2516 g, the percentage by weight of CaCO₃ is:

$\frac{1,168g}{1,2516g}$×100 = 93,32 % (w/w)

I hope it helps!