A 1.06 gram sample of limestone (mainly calcium carbonate) was dissolved in an a

Question

A 1.06 gram sample of limestone (mainly calcium carbonate) was dissolved in an acidic solution, and then the calcium was precipitated as insoluble calcium oxalate (CaC2O4). The CaC2O4 precipitate is filtered, washed and is then dissolved in acid. The resulting solution was treated with KMnO4 (Potassium permanganate). It was found that 0.694 grams of KMnO4 (mol wt = 158.0) were required to react completely with the solution. Determine the percentage of Ca (at wt = 40.1) in the original sample.

Hint: the unbalanced redox equation involved is:

(C2O4)^-2 + (MnO4)^-1 ------> CO2 + (Mn)^+2

Explanation / Answer

balanced equation:

5 (C2O4)2- + 2 (MnO4)2- -----> 10CO2 + 2Mn2+

Moles of KmNO4 = 0.694 / 158 = 0.004392

Moles of (C2O4)2- = moles o KMnO4 * 5/2 = 0.01098

moles of calcium = 0.01098 * 2 = 0.02196

mass of calcium = 0.02196 * 40.1 = 0.88067

% of Calcium = (0.88067 / 1.06) * 100

= 83.08%

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