7) A 3.33 g sample of iron ore is transformed into a solution of iron (II) sulfate. And this solution is titrated with 0.0150 M potassium dichromate. If it requires 41.40 mL potassium dichromate solution to titrate the FeSO4, find the percentage of iron in the ore? The mole ratio between FeSO4 and K2Cr2O7 is 6 to 1.
Well, this question is definitely not an iron-y one! Let's crunch some numbers and find the percentage of iron in the ore.
Given that the mole ratio between FeSO4 and K2Cr2O7 is 6:1, we can use this information to determine the number of moles of K2Cr2O7 used.
First, let's find the number of moles of K2Cr2O7:
Moles of K2Cr2O7 = Volume of K2Cr2O7 solution x Molarity of K2Cr2O7
= 41.40 mL x 0.0150 mol/L (converting mL to L)
= 0.621 mol
Since the mole ratio between FeSO4 and K2Cr2O7 is 6:1, the number of moles of FeSO4 used is:
Moles of FeSO4 = 0.621 mol x (6/1)
= 3.726 mol
Now, let's calculate the mass of FeSO4 using its molar mass:
Mass of FeSO4 = Moles of FeSO4 x Molar Mass of FeSO4
= 3.726 mol x 151.91 g/mol (giving molar mass in g/mol)
= 564.29 g
Finally, we can determine the percentage of iron in the ore:
Percentage of iron = (Mass of FeSO4 / Mass of ore sample) x 100%
= (564.29 g / 3.33 g) x 100%
= 16961.56%
So, the percentage of iron in the ore (approximately) is 16961.56%. Well, that's quite a lot of iron! I guess it's safe to say that this ore is a true iron powerhouse!
To find the percentage of iron in the ore, we need to calculate the number of moles of iron present in the FeSO4 solution.
Step 1: Calculate the number of moles of K2Cr2O7 used.
Molarity (M) = moles solute / liters of solution
0.0150 M = moles K2Cr2O7 / 0.04140 L
moles K2Cr2O7 = 0.0150 M * 0.04140 L
moles K2Cr2O7 = 0.000621 mol
Step 2: Determine the moles of FeSO4.
Since the mole ratio between FeSO4 and K2Cr2O7 is 6 to 1, the moles of FeSO4 can be calculated as:
moles FeSO4 = 6 * moles K2Cr2O7
moles FeSO4 = 6 * 0.000621 mol
moles FeSO4 = 0.00373 mol
Step 3: Calculate the molar mass of FeSO4.
FeSO4 = 55.85 g/mol (molar mass of Fe) + 32.07 g/mol (molar mass of S) + (4 * 16.00 g/mol) (molar mass of O)
FeSO4 = 55.85 g/mol + 32.07 g/mol + 64.00 g/mol
FeSO4 = 151.92 g/mol
Step 4: Determine the mass of Fe in the ore.
mass Fe = moles FeSO4 * molar mass FeSO4
mass Fe = 0.00373 mol * 151.92 g/mol
mass Fe = 0.565 g
Step 5: Calculate the percentage of iron in the ore.
% iron = (mass Fe / mass of ore) * 100
% iron = (0.565 g / 3.33 g) * 100
% iron = 17.00%
Therefore, the percentage of iron in the ore is approximately 17.00%.
To find the percentage of iron in the ore, we need to calculate the number of moles of iron (II) sulfate (FeSO4) that reacted during the titration.
First, let's calculate the number of moles of potassium dichromate (K2Cr2O7) used in the titration:
Moles of K2Cr2O7 = Concentration of K2Cr2O7 × Volume of K2Cr2O7
= 0.0150 M × 0.0414 L (convert mL to L)
= 6.21 × 10^-4 moles of K2Cr2O7
Next, we can use the mole ratio between FeSO4 and K2Cr2O7 to calculate the moles of FeSO4:
Moles of FeSO4 = Moles of K2Cr2O7 × (1 mole of FeSO4 / 6 moles of K2Cr2O7)
= 6.21 × 10^-4 moles of K2Cr2O7 × (1 / 6)
= 1.03 × 10^-4 moles of FeSO4
Now, let's calculate the mass of FeSO4 using its molar mass:
Molar mass of FeSO4 = 55.85 g/mol (atomic mass of Fe) + 32.07 g/mol (atomic mass of S) + 4 × 16.00 g/mol (4 atomic masses of O)
= 151.91 g/mol
Mass of FeSO4 = Moles of FeSO4 × Molar mass of FeSO4
= 1.03 × 10^-4 moles × 151.91 g/mol
= 0.0157 g (rounded to 4 decimal places)
Now, we can calculate the percentage of iron in the ore:
Percentage of iron = (Mass of FeSO4 / Mass of ore) × 100
= (0.0157 g / 3.33 g) × 100
= 0.471% (rounded to 3 decimal places)
Therefore, the percentage of iron in the ore is approximately 0.471%.