Solud iron and steam react at high temperatures to form iron(vi) peroxide and hydrogen gas. How much solid iron is needed to obtained 405 grams of iron(vi) peroxide, with a 84.2% yield?

To determine the amount of solid iron needed to obtain 405 grams of iron(VI) peroxide with an 84.2% yield, we need to use stoichiometry and the concept of percent yield.

First, let's balance the chemical equation for the reaction between solid iron (Fe) and steam (H2O):
3Fe + 4H2O → Fe3O4 + 4H2

From the balanced equation, we can see that 3 moles of iron react with 4 moles of water to produce 1 mole of iron(VI) peroxide (Fe3O4) and 4 moles of hydrogen gas (H2).

Step 1: Calculate the molecular weight of Fe3O4.
The molecular weight of Fe = 55.845 g/mol
The molecular weight of O = 16.00 g/mol
Since Fe3O4 consists of three iron atoms and four oxygen atoms, the molecular weight of Fe3O4 can be calculated as:
(3 x Fe) + (4 x O) = (3 x 55.845 g/mol) + (4 x 16.00 g/mol) = 231.687 g/mol

Step 2: Determine the number of moles of Fe3O4.
We know that the mass of Fe3O4 is 405 grams. To convert this mass to moles, we divide the mass by the molecular weight:
moles of Fe3O4 = mass of Fe3O4 / molecular weight of Fe3O4
moles of Fe3O4 = 405 g / 231.687 g/mol

Step 3: Calculate the theoretical yield of Fe3O4.
The theoretical yield is the maximum amount of product that can be obtained in a reaction if it goes to completion. Since the balanced equation shows that 3 moles of Fe react to produce 1 mole of Fe3O4, we can use the moles of Fe3O4 to calculate the moles of Fe needed:
moles of Fe = (moles of Fe3O4) x (3 moles of Fe / 1 mole of Fe3O4)

Step 4: Determine the actual yield using percent yield.
The percent yield is a measure of how efficient the reaction is and is given as a percentage. In this case, the percent yield is 84.2%. To determine the actual yield, we multiply the theoretical yield by the percent yield:
actual yield = theoretical yield x (percent yield / 100)

Step 5: Calculate the mass of solid iron needed.
To calculate the mass of solid iron needed, we can use the moles of Fe and the molecular weight of Fe:
mass of Fe = moles of Fe x molecular weight of Fe

By following these steps, you can determine the amount of solid iron needed to obtain 405 grams of iron(VI) peroxide with an 84.2% yield.

To calculate the amount of solid iron needed, we first need to determine the molar mass of iron(VI) peroxide (FeO3).

The molar mass of FeO3 can be calculated using the periodic table:
- The molar mass of iron (Fe) is 55.845 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol, and there are three oxygen atoms in FeO3.

So the molar mass of FeO3 is: (1 * 55.845) + (3 * 16.00) = 159.845 g/mol.

Now we can calculate the theoretical yield of FeO3 using the molar ratio between FeO3 and Fe. From the balanced equation, we know that 1 mole of FeO3 is produced from 1 mole of Fe.

Given that the yield is 84.2%, the actual yield is: (84.2/100) * 405 g = 341.61 g.

Using the molar mass of FeO3, the number of moles of FeO3 produced is: 341.61 g / 159.845 g/mol = 2.141 moles.

Since 1 mole of FeO3 is produced from 1 mole of Fe, the number of moles of Fe needed is also 2.141 moles.

Finally, we can calculate the mass of solid iron needed using the molar mass of iron (Fe):
Mass of solid iron = Number of moles of Fe * Molar mass of Fe
= 2.141 moles * 55.845 g/mol
= 119.69 g

Therefore, approximately 119.69 grams of solid iron is needed to obtain 405 grams of iron(VI) peroxide, with a 84.2% yield.