Question

A 1.2516 gram sample of a mixture of caco3 and na2so4 was analyzed by dissolving the sample and completely precipitating the ca as cac2o4.the cac2o4 was dissolved in sulfuric acid and the resulting h2c2o4 was titrated with a standard kmno4 solution. the titration required 35.62 milliliters of 0.1092 m kmno4.calculate the number the number of moles of h2c2o4 that reacted with the kmno4.calculate the number of moles of caco3 in the original sample. calculate the percentage by weight of caco3 in the original sample.

Answer 1

The number of moles of H₂C₂O₄ that reacted with KMnO₄ is 0.003889464 mol. From this, the moles of CaCO₃ in the original sample is also 0.003889464 mol. The percentage by weight of CaCO₃ in the mixture is approximately 31.15%.

Step-by-step explanation:

The number of moles of H₂C₂O₄ that reacted with the KMnO₄ is calculated using the molarity and volume of KMnO₄ solution used in the titration. Using the balanced chemical equation for the reaction between H₂C₂O₄ and KMnO₄, which is MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O, we can find the corresponding moles of CaCO₃ in the original sample. Finally, the percentage by weight (%w/w) of CaCO₃ in the original sample can be calculated by dividing the mass of CaCO₃ obtained from its moles by the initial mass of the sample and multiplying by 100.

To calculate the moles of H₂C₂O₄, we use the titration data:
# moles H₂C₂O₄ = (0.03562 L)(0.1092 M) = 0.003889464 mol H₂C₂O₄.

To find the moles of CaCO₃, we consider that 1 mole of CaCO₃ produces 1 mole of H₂C₂O₄:
# moles CaCO₃ = # moles H₂C₂O₄ = 0.003889464 mol.

To find the %w/w of CaCO₃:
%w/w CaCO₃ = (moles CaCO₃ × molar mass CaCO₃ ÷ initial mass of the sample) × 100 = (0.003889464 mol × 100.087 g/mol ÷ 1.2516 g) × 100 ≈ 31.15%.

Answer 2

0.009725 moles of H2C2O4

0.009725 moles CaCO3

Mass percentage = 77.77%

Step-by-step explanation:

Step 1: The balanced equation

2MnO4- +5C2H2O4+6H+ →2Mn2+ +10CO2+8H2O

We can see that for 2 moles of Mno4- consumed , there is 5 moles of C2H2O4 needed and 6 moles H+ to produce 2 moles Mn2+, 10 moles of CO2 and 8 moles of H2O

Step 2: Calculate moles of MnO4-

Molarity = Moles/volume

Moles of Mno4- = Molarity of MnO4- * Volume of Mno4-

Moles of Mno4- = 0.1092M * 35.62 *10^-3 L

Moles of MnO4- = 0.00389 moles

Step 3: Calculate moles of H2C2O4

Since there is needed 5 moles of C2H2O4 to consume 2 moles of MnO4-

then for 0.00389 moles of MnO4-, there is 5/2 *0.00389 = 0.009725 moles of H2C2O4

Step 4: Calculate moles of CaCO3

moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3

Step 5: Calculate mass of CaCO3

Molar mass of CaCO3 = 100.09 g/mole

Mass of CaCO3 = moles of CaCO3 * Molar mass of CaCO3

Mass of CaCO3 = 0.009725 moles * 100.09 g/mole = 0.9734 g

Step 6: Calculate percentage by weight of CaCO3

Mass of CaCO3 = 0.9734g

Mass of original sample = 1.2516g

Mass percentage = 0.9734/1.2516 *100% = 77.77%