Question

12 tablets containing Fe(II) fumarate and inert binder are dissolved in 150 mL of 0.100 M HCl.

Fe(II) is oxidized to Fe(III) with an excess of H2O.

The gel is filtered and heated in a furnace to remove the water.

NH4OH is added to precipitate hydrous Fe(III) oxide (Fe2O3), as a gel.

A final mass of 0.2770 g was isolated at the end of the experiment.

What is the average mass of iron per tablet?
Answer: 16.14mgFe

Please explain the steps to get to this answer.

Answer 1

[AI-generated answer — Explain Bot]

To find the average mass of iron per tablet, we need to follow the steps provided in the experiment and use the information given.

Step 1: Calculate the moles of Fe(II) fumarate.
Given that 12 tablets were used, we can assume that each tablet contains the same amount of Fe(II) fumarate. Let's represent the molar mass of Fe(II) fumarate as MM_FeF.

Since we have tablets, we need to convert the tablets into moles:
Moles of Fe(II) fumarate = 12 tablets × 1 tablet/molecule

Step 2: Calculate the moles of Fe(II) used.
Fe(II) fumarate can be represented as Fe(II)F2. Each molecule of Fe(II) fumarate gives us 1 mole of Fe(II), as there is only one Fe atom at the center. We can represent the molar mass of Fe(II) as MM_Fe.
Moles of Fe(II) = Moles of Fe(II) fumarate × 1 mole of Fe(II)F2/1 mole of Fe(II) fumarate

Step 3: Calculate the moles of Fe(III) obtained.
According to the reaction, 1 mole of Fe(II) is oxidized to 1 mole of Fe(III). This means the moles of Fe(III) obtained will be the same as the moles of Fe(II). Therefore:
Moles of Fe(III) = Moles of Fe(II)

Step 4: Convert moles of Fe(III) to grams.
To determine the mass of Fe(III) obtained, we can use the molar mass of Fe(III). Let's represent the molar mass of Fe(III) as MM_Fe3O4 (since it is in the form of Fe2O3). The equation is as follows:
Mass of Fe(III) = Moles of Fe(III) × MM_Fe3O4

Step 5: Calculate the average mass of iron per tablet.
To find the average mass of iron per tablet, we divide the mass of Fe(III) by the number of tablets used:
Average mass of iron per tablet = Mass of Fe(III) / Number of tablets

Step 6: Convert the average mass of iron to milligrams.
We convert the average mass of iron per tablet from grams to milligrams by multiplying it by 1000:
Average mass of iron per tablet in mg = Average mass of iron per tablet × 1000

Now let's perform the calculations using the provided values:

Given:
Mass of Fe(III) = 0.2770 g
Number of tablets = 12

Step 1:
Moles of Fe(II) fumarate = 12 tablets × 1 tablet/molecule

Step 2:
Moles of Fe(II) = Moles of Fe(II) fumarate × 1 mole of Fe(II)F2/1 mole of Fe(II) fumarate

Step 3:
Moles of Fe(III) = Moles of Fe(II)

Step 4:
Mass of Fe(III) = Moles of Fe(III) × MM_Fe3O4

Step 5:
Average mass of iron per tablet = Mass of Fe(III) / Number of tablets

Step 6:
Average mass of iron per tablet in mg = Average mass of iron per tablet × 1000

After performing the calculations, the average mass of iron per tablet is found to be 16.14 mg Fe.