Based on the equation of the reaction;
- the mass of iron (ii) ion present in the sample ore is 0.31 g
- the percent mass of iron in the ore is 22.8%
<h3>What is the percentage mass of iron in the ore?</h3>
The percentage mass of iron in the ore is determined from the net ionic equation of the reaction as follows:
5 Fe²⁺ + MnO₄⁻ + 8H+ ---> 5 Fe³⁺ + Mn²⁺ + 4 H₂O
The mole ratio of the reaction shows that 5 moles of the iron (ii) ion is oxidized by 1 mole of tetraoxomanganate (vii) ion, MnO₄⁻.
The moles of tetraoxomanganate (vii) ion, MnO₄⁻ that reacted with al the iron present in the ore is determined as follows:
Moles of substance = molarity * volume in liters
Moles of tetraoxomanganate (vii) ion, MnO₄⁻ = 0.0281 * 39.42/1000
Moles of tetraoxomanganate (vii) ion, MnO₄⁻ = 0.001108 moles
Moles of iron (ii) ion present in the sample = 0.001108 * 5
Moles of iron (ii) ion present in the sample = 0.00554 moles
Mass of iron (ii) ion present in the sample = 0.00554 * 56
Mass of iron (ii) ion present in the sample = 0.31 g
Percent mass of iron in the ore = 0.31/1.362 * 100%
Percent mass of iron in the ore = 22.8%
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