1.630 g of iron ore is dissolved in an acidic solution. This solution is titrated to a pink endpoint with 27.15 mL of a 0.020 M KMnO4 solution.

a. How many moles of MnO4- ions were consumed?

b. How many moles of Fe2+ were in the iron ore sample?

c. What is the percent of iron in the iron ore sample?

a. Well, this is quite a colorful situation we have here! To figure out how many moles of MnO4- ions were consumed, we need to use the molarity of the KMnO4 solution and the volume of the solution used. So, let's do some calculations.

First, convert the volume of the KMnO4 solution into liters: 27.15 mL = 0.02715 L.

Next, we can use the formula Molarity (M) = moles (mol) / volume (L) to find the moles of MnO4- ions consumed. Rearranging the formula gives us moles = Molarity x volume.

Moles of MnO4- ions consumed = 0.020 M x 0.02715 L = 0.000543 mol.

So, approximately 0.000543 moles of MnO4- ions were consumed.

b. Now, let's talk about our friendly Fe2+ ions hiding in the iron ore. To find the moles of Fe2+ in the sample, we need to use the balanced equation between Fe2+ and MnO4-.

From the balanced equation, we know that 1 mole of MnO4- reacts with 5 moles of Fe2+. Since we know the moles of MnO4- consumed (0.000543 mol), we can determine the moles of Fe2+ using this ratio.

Moles of Fe2+ = 5 x 0.000543 mol = 0.002715 mol.

So, approximately 0.002715 moles of Fe2+ were in the iron ore sample.

c. Ah, the percent of iron in the iron ore sample! To find this, we need to use the molar mass of iron and the mass of the iron ore sample dissolved.

The molar mass of iron (Fe) is approximately 55.85 g/mol. And we know the mass of the iron ore sample dissolved is 1.630 g.

Now, let's calculate the percent of iron in the iron ore sample:

Percent of iron = (moles of Fe2+ x molar mass of Fe) / mass of iron ore sample x 100%.

Percent of iron = (0.002715 mol x 55.85 g/mol) / 1.630 g x 100%.

After some math magic, we find that the percent of iron in the iron ore sample is approximately 8.37%.

So there you have it, the iron ore sample has a dash of humor and approximately 8.37% iron in it!

a. To find the number of moles of MnO4- ions consumed, we can use the balanced equation of the reaction between the MnO4- ions and Fe2+ ions. From the equation, we can see that 1 mole of MnO4- ions reacts with 5 moles of Fe2+ ions.

The balanced equation is:

5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O

From the question, we know that 27.15 mL of a 0.020 M KMnO4 solution is used. To convert this to moles, we can use the formula:

moles of solute = concentration (M) x volume (L)

So, the moles of MnO4- ions consumed can be calculated as:

moles of MnO4- = 0.020 M x 0.02715 L = 0.000543 moles

b. Since 1 mole of MnO4- ions reacts with 5 moles of Fe2+ ions, the moles of Fe2+ ions can be calculated as:

moles of Fe2+ = (moles of MnO4- ions consumed) / 5 = 0.000543 moles / 5 = 0.000109 moles

c. To find the percent of iron in the iron ore sample, we need to know the molar mass of iron (Fe). The molar mass of Fe is approximately 55.845 g/mol.

The mass of iron in the sample can be calculated as:

mass of Fe = (moles of Fe2+) x (molar mass of Fe) = 0.000109 moles x 55.845 g/mol = 0.00609 g

The percent of iron can be calculated as:

percent of iron = (mass of Fe / mass of iron ore sample) x 100 = (0.00609 g / 1.630 g) x 100 = 0.374% (rounded to three decimal places)

Therefore, the percent of iron in the iron ore sample is approximately 0.374 %.

To solve these questions, we will use the concept of stoichiometry, which involves calculating the amount of reactants and products in a chemical reaction. In this case, we are given the mass of iron ore, the volume and concentration of the KMnO4 solution, and we need to find the number of moles and the percent of iron in the sample.

a. First, let's calculate the number of moles of MnO4- ions consumed:

1. Convert the volume of the KMnO4 solution from milliliters to liters:
27.15 mL * (1 L / 1000 mL) = 0.02715 L

2. Calculate the number of moles of MnO4- ions consumed using the formula:
Moles = Volume (L) * Concentration (mol/L)
Moles = 0.02715 L * 0.020 mol/L = 0.000543 mol

b. Next, we need to find the number of moles of Fe2+ in the iron ore sample. The balanced equation for the reaction between Fe2+ and MnO4- is:

5 Fe2+ + MnO4- + 8 H+ -> 5 Fe3+ + Mn2+ + 4 H2O

From the equation, we can see that each mole of MnO4- reacts with 5 moles of Fe2+ ions. Therefore, the number of moles of Fe2+ in the sample will be 5 times the number of moles of MnO4- consumed:

Moles of Fe2+ = 5 * 0.000543 = 0.002715 mol

c. Finally, we can calculate the percent of iron in the iron ore sample. To do this, we need to know the molar mass of iron. The molar mass of iron is approximately 55.85 g/mol.

The percent of iron in the iron ore sample can be calculated using the formula:
Percent of iron = (Moles of Fe2+ * Molar mass of iron * 100) / Mass of iron ore sample

Given that the mass of iron ore sample is 1.630 g, we can now substitute the values into the formula:

Percent of iron = (0.002715 mol * 55.85 g/mol * 100) / 1.630 g

Using a calculator, this comes out to be approximately 86.51%.

Therefore:
a. The number of moles of MnO4- ions consumed is 0.000543 mol.
b. The number of moles of Fe2+ in the iron ore sample is 0.002715 mol.
c. The percent of iron in the iron ore sample is approximately 86.51%.

a. mols KMnO4 = M x L = about 0.00054 but you need to do that more accurately. Thjat's just an estimate.

b. 5Fe^2+ + MnO4^- ==> 5Fe^3+ + Mn^2+
Note that I balanced only the redox part to save a little time but that's all we really need to balance.
So mols Fe^2+ must be 5*mols MnO4^- = about 0.0027 mols Fe.

c. Convert mols Fe to grams Fe.
mols Fe x atomic mass Fe = approx 0.15
Then % = (g Fe/mass sample)*100 = ?

c.