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katrin [286]
1 year ago
12

A 1.362 g sample of an iron ore that contained Fe3O4 was dissolved in acid and all of the iron was reduced to Fe2+. The solution

was then acidified with H2SO4 and titrated with 39.42 mL of 0.0281 M KMnO4, which oxidized the iron to Fe3+. The net ionic equation for the reaction is 5Fe2+ + MnO4- + 8H+ h 5Fe3+ + Mn2+ + 4H2O (a) What was the percentage by mass of iron in the ore? (b) What was the percentage by mass of Fe3O4 in the ore?
Chemistry
1 answer:
tino4ka555 [31]1 year ago
4 0

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%

Learn more about percentage mass at: brainly.com/question/26150306

#SPJ1

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3 0
3 years ago
How many molecules of CaCl2 are equivalent to 75.9 g CaCl2
sergij07 [2.7K]
First, you need to find:
One mole of CaCl_{2} is equivalent to how many grams?

Well, for this you have to look up the periodic table. According to the periodic table:
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As there are two atoms of Chlorine present in CaCl_{2}, therefore, the atomic mass of CaCl_{2} would be:

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Atomic mass of CaCl_{2} = 40.078 + 2 * 35.45 = 110.978 g

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110.978 g of CaCl_{2} = 1 mole.
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According to <span>Avogadro's number,
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