A 3.75-G SAMPLE OF IRON ORE IS TRANSFORMED TO A SOLUTION OF IRON(II)SULFATE, FeSO4,AND THIS SOLUTION IS TITRATED WITH 0.150 M K2Cr2O7. IF IT REQUIRES 43.7 mL OF POTASSIUM DICHROMATE SOLUTION TO TITRATE THE IRON(II)SULFATE SOLUTION, WHAT IS THE PERCENTAGE OF IRON IN THE ORE? THE REACTION IS

6FeSO4(aq)+K2Cr207(aq)+7H2SO4(aq)�¨
3Fe2(SO4)3(aq)+Cr2(SO4)3(aq)+7H2O(l)+K2SO4(aq)

mols dichromate = M x L = ?

Use the coefficients in the balanced equation to convert mols dichromate to mols FeSO4
Convert mols FeSO4 to mols Fe then to grams Fe. g Fe = mols Fe x atomic mass Fe.
%Fe = (mass Fe/mass sample)*100 = ?

62.81%

To calculate the percentage of iron in the ore, we first need to determine the moles of iron reacted with the potassium dichromate.

1. Calculate the moles of K2Cr2O7:
- Given the volume of K2Cr2O7 solution (43.7 mL) and its concentration (0.150 M), we can calculate the moles of K2Cr2O7 using the formula:
Moles of K2Cr2O7 = Volume (L) × Concentration (mol/L)
=> Moles of K2Cr2O7 = 43.7 mL × 0.150 M / 1000 mL/L

2. Use the balanced equation to determine the moles of iron reacted:
- From the balanced equation, we can see that the stoichiometric ratio between FeSO4 and K2Cr2O7 is 6:1.
- Therefore, the moles of FeSO4 reacted will be 6 times the moles of K2Cr2O7.

3. Calculate the moles of FeSO4:
- Moles of FeSO4 = Moles of K2Cr2O7 × 6

4. Convert the moles of FeSO4 to grams:
- The molar mass of FeSO4 is:
Fe = 55.845 g/mol
S = 32.06 g/mol
O = 16.00 g/mol (x4 from the formula)
=> Molar mass of FeSO4 = 55.845 + 32.06 + 16.00(4) g/mol

- Mass of FeSO4 = Moles of FeSO4 × Molar mass of FeSO4

5. Calculate the percentage of iron in the ore:
- Percentage of iron = (Mass of FeSO4 / Mass of ore) × 100%

Remember to convert mL to L in the calculations.

To find the percentage of iron in the ore, we need to calculate the amount of iron(II) sulfate formed during the reaction and then convert it to a percentage.

Here is the step-by-step process:

1. Calculate the number of moles of potassium dichromate (K2Cr2O7) used in the titration:

Moles of K2Cr2O7 = concentration of K2Cr2O7 × volume of K2Cr2O7 solution used
= 0.150 M × 43.7 mL (convert mL to L by dividing by 1000)
= 0.150 M × 0.0437 L
= 0.006555 moles of K2Cr2O7

2. From the balanced chemical equation, we can see that the stoichiometric ratio between K2Cr2O7 and FeSO4 is 1:6. So, for every mole of K2Cr2O7, we have 6 moles of FeSO4 reacted.

Moles of FeSO4 = 6 × moles of K2Cr2O7
= 6 × 0.006555 moles
= 0.039330 moles of FeSO4

3. Calculate the molar mass of FeSO4:

Fe: 55.85 g/mol
S: 32.07 g/mol
O: 3 × 16.00 g/mol = 48.00 g/mol

Molar mass of FeSO4 = 55.85 + 32.07 + 48.00 = 135.92 g/mol

4. Calculate the mass of FeSO4 formed:

Mass of FeSO4 = moles of FeSO4 × molar mass of FeSO4
= 0.039330 moles × 135.92 g/mol
= 5.340 g of FeSO4

5. Finally, calculate the percentage of iron in the ore:

Percentage of iron = (mass of FeSO4 / mass of ore sample) × 100%
= (5.340 g / 3.75 g) × 100%
= 142.4%

Therefore, the percentage of iron in the ore is approximately 142.4%.