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Question 7 Pyrolusite, an ore of manganese, contains manganese in the form of MnO₂ - VCE - SSCE Chemistry - Question 7 - 2003 - Paper 1

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Question 7 Pyrolusite, an ore of manganese, contains manganese in the form of MnO₂. A sample of pyrolusite from a newly discovered deposit is analysed to determine t... show full transcript

Worked Solution & Example Answer:Question 7 Pyrolusite, an ore of manganese, contains manganese in the form of MnO₂ - VCE - SSCE Chemistry - Question 7 - 2003 - Paper 1

Step 1

Calculate the amount in mole of oxalic acid remaining in the original 100 mL solution after the pyrolusite had been reacted with the oxalic acid.

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Answer

  1. Calculate the initial moles of oxalic acid: n(H2C2O4)=C×V=0.150 mol/L×0.100 L=0.0150 moln(H_2C_2O_4) = C \times V = 0.150 \, mol/L \times 0.100 \, L = 0.0150 \, mol

  2. Calculate the moles of oxalic acid reacted with I₃⁻: n(I_3^-) = C \times V = 0.050 , mol/L \times 0.022 , L = 0.00110 , mol

  3. The stoichiometry for the reaction shows: n(H2C2O4)reacted=2×n(I3−)=2×0.00110=0.00220 moln(H_2C_2O_4)_{reacted} = 2 \times n(I_3^-) = 2 \times 0.00110 = 0.00220 \, mol

  4. Remaining moles of oxalic acid: n(H2C2O4)remaining=n(H2C2O4)initial−n(H2C2O4)reactedn(H_2C_2O_4)_{remaining} = n(H_2C_2O_4)_{initial} - n(H_2C_2O_4)_{reacted} =0.0150−0.00220=0.01280 mol= 0.0150 - 0.00220 = 0.01280 \, mol

Step 2

Calculate the amount in mole of oxalic acid used to reduce the MnO₂ in 1.25 g of pyrolusite.

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Answer

  1. The moles of oxalic acid used is: n(H2C2O4)used=n(H2C2O4)initial−n(H2C2O4)remainingn(H_2C_2O_4)_{used} = n(H_2C_2O_4)_{initial} - n(H_2C_2O_4)_{remaining} =0.0150−0.01280=0.00220 mol= 0.0150 - 0.01280 = 0.00220 \, mol

Step 3

Calculate the amount in mole of MnO₂ present in the original 1.25 g of pyrolusite and hence the percentage of MnO₂ by mass present in the pyrolusite.

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Answer

  1. From the reaction, the stoichiometry shows: 1 mol  MnO2  reacts  with  1 mol  H2C2O41 \, mol \; MnO_2 \; reacts \; with \; 1 \, mol \; H_2C_2O_4

  2. Therefore, the moles of MnO₂ is equal to the moles of oxalic acid used: n(MnO2)=n(H2C2O4)used=0.00220 moln(MnO_2) = n(H_2C_2O_4)_{used} = 0.00220 \, mol

  3. Calculate the mass of MnO₂: mass=n×molar  mass=0.00220×86.9=0.19118 gmass = n \times molar \; mass = 0.00220 \times 86.9 = 0.19118 \, g

  4. Calculate the percentage of MnO₂ by mass in pyrolusite: percentage  of  MnO2=(mass  of  MnO2mass  of  pyrolusite)×100percentage \; of \; MnO_2 = \left( \frac{mass \; of \; MnO_2}{mass \; of \; pyrolusite} \right) \times 100 =(0.191181.25)×100=15.29%= \left( \frac{0.19118}{1.25} \right) \times 100 = 15.29 \%

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