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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

To find the amount of oxalic acid remaining, we first need to calculate the initial moles of oxalic acid in the 100 mL solution:

n(H2C2O4)=CimesV=0.150imes0.100=0.0150extmoln(H₂C₂O₄) = C imes V = 0.150 imes 0.100 = 0.0150 ext{ mol}

Next, we determine the moles of oxalic acid that reacted. The reaction with the triodide ion can be used to find the moles of oxalic acid consumed:

n(I3−)=CimesV=0.0510imes0.022=0.00112extmoln(I₃⁻) = C imes V = 0.0510 imes 0.022 = 0.00112 ext{ mol}

From the stoichiometry of the reaction: 1 mole of oxalic acid reacts with 1 mole of I₃⁻. Thus, 0.00112 mol of oxalic acid was consumed.

Now, we can calculate the remaining amount:

n(H2C2O4)extremaining=n(H2C2O4)extinitial−n(H2C2O4)extusedn(H₂C₂O₄) ext{ remaining} = n(H₂C₂O₄) ext{ initial} - n(H₂C₂O₄) ext{ used}

=0.0150−0.00112=0.01388extmol= 0.0150 - 0.00112 = 0.01388 ext{ 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

To calculate the moles of oxalic acid used to reduce the MnO₂, we first need to know how much remained. We calculated that 0.01388 mol of oxalic acid remained.

To find out how much was used:

n(H2C2O4)extused=n(H2C2O4)extinitial−n(H2C2O4)extremainingn(H₂C₂O₄) ext{ used} = n(H₂C₂O₄) ext{ initial} - n(H₂C₂O₄) ext{ remaining}

=0.0150−0.01388=0.00112extmol= 0.0150 - 0.01388 = 0.00112 ext{ 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

The moles of MnO₂ can be found using the stoichiometry of the reaction with oxalic acid. From the balanced equation:

MnO2+H2C2O4→Mn2++2CO2+2H2OMnO₂ + H₂C₂O₄ → Mn^{2+} + 2CO₂ + 2H₂O

It shows that 1 mole of MnO₂ reacts with 1 mole of oxalic acid. Thus, the moles of MnO₂ reacting is:

n(MnO2)=n(H2C2O4)extused=0.00112extmoln(MnO₂) = n(H₂C₂O₄) ext{ used} = 0.00112 ext{ mol}

To find the mass of MnO₂, we use its molar mass ( ext{approximately} 86.94 g/mol):

m=nimesM=0.00112imes86.94=0.0974extgm = n imes M = 0.00112 imes 86.94 = 0.0974 ext{ g}

The percentage by mass of MnO₂ in the pyrolusite sample can be calculated as follows:

ext{Percentage of} MnO₂ = rac{m(MnO₂)}{m( ext{pyrolusite})} imes 100 ext{ } = rac{0.0974}{1.25} imes 100 = 7.79 ext{ }\%

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