the balanced equation for the reduction of iron ore to the metal using CO is Fe2O3+3CO(g) -> 2Fe+3CO so what is the maximum of iron in grams that can bbe obtained from 454g (1.00 lb) of iron (III) oxide? AND what mass of CO is required to react with the 454g of Fe2O3? please explain each step in detail

This link will solve any stoichiometry problem you have. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

Very helpful

Why did the iron ore start going to therapy?

Because it couldn't handle all that pressure!

Now, let's answer your questions in a step-by-step manner:

Step 1: Calculate the molar mass of Fe2O3.
The molar mass of iron (III) oxide (Fe2O3) can be calculated by adding the atomic masses of its constituent elements:
2(Fe) + 3(O) = 2(55.845 g/mol) + 3(16.00 g/mol) = 159.69 g/mol

Step 2: Calculate the number of moles of Fe2O3 in 454g.
To find the number of moles, divide the mass of Fe2O3 by its molar mass:
454 g / 159.69 g/mol ≈ 2.85 mol

Step 3: Determine the limiting reactant.
To do this, we compare the stoichiometric ratios of the reactants in the balanced equation. The ratio of Fe2O3 to CO is 1:3. Based on this ratio, we can see that 1 mole of Fe2O3 requires 3 moles of CO to react completely. Therefore, the limiting reactant is CO, as it is present in a smaller stoichiometric amount.

Step 4: Calculate the moles of CO required.
Since CO is the limiting reactant, the number of moles of CO required can be determined using the stoichiometry of the balanced equation. From the equation, 3 moles of CO are required for every 1 mole of Fe2O3:
2.85 mol Fe2O3 × (3 mol CO / 1 mol Fe2O3) = 8.55 mol CO

Step 5: Convert moles of CO to grams.
To convert moles of CO to grams, we use the molar mass of CO, which is calculated by adding the atomic masses of carbon and oxygen:
12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Now, calculate the mass of CO:
8.55 mol CO × 28.01 g/mol ≈ 239.70 g

So, to summarize:
- The maximum mass of iron that can be obtained from 454g of Fe2O3 is the mass of Fe produced by the complete reaction. According to the stoichiometry of the equation, 1 mole of Fe2O3 produces 2 moles of Fe. Thus, 2.85 mol of Fe2O3 will produce 2.85 × 2 = 5.70 mol of Fe. Finally, using the atomic mass of Fe (which is 55.845 g/mol), we find that the mass of Fe produced is 5.70 mol × 55.845 g/mol ≈ 318.18 g.
- The mass of CO required to react with 454g of Fe2O3 is approximately 239.70 g.

Remember, chemistry can be quite reactive, so handle those equations with care!

To find the maximum amount of iron in grams that can be obtained from 454g (1.00 lb) of iron (III) oxide (Fe2O3), we need to use stoichiometry.

Step 1: Calculate the molar mass of Fe2O3.
Fe = 55.85 g/mol
O = 16.00 g/mol
Molar mass of Fe2O3 = (2 * 55.85) + (3 * 16.00) = 159.70 g/mol

Step 2: Calculate the number of moles of Fe2O3.
Number of moles = mass (g) / molar mass (g/mol)
Number of moles = 454g / 159.70 g/mol ≈ 2.845 mol

Step 3: Use the balanced equation to determine the stoichiometry ratio.
According to the balanced equation: Fe2O3 + 3CO(g) -> 2Fe + 3CO

The stoichiometry ratio of Fe2O3 to Fe is 1:2, which means that 1 mole of Fe2O3 will yield 2 moles of Fe.

Step 4: Calculate the number of moles of Fe.
Number of moles of Fe = Number of moles of Fe2O3 * stoichiometry ratio
Number of moles of Fe = 2.845 mol * 2 = 5.69 mol

Step 5: Calculate the mass of Fe.
Mass of Fe = Number of moles of Fe * molar mass of Fe
Mass of Fe = 5.69 mol * 55.85 g/mol = 317.59 g

Therefore, the maximum amount of iron that can be obtained from 454g (1.00 lb) of iron (III) oxide (Fe2O3) is approximately 317.59 grams.

Next, let's calculate the mass of CO required to react with the 454g of Fe2O3.

Step 1: Calculate the molar mass of CO.
C = 12.01 g/mol
O = 16.00 g/mol
Molar mass of CO = 12.01 + 16.00 = 28.01 g/mol

Step 2: Use the balanced equation to determine the stoichiometry ratio.
According to the balanced equation: Fe2O3 + 3CO(g) -> 2Fe + 3CO

The stoichiometry ratio of Fe2O3 to CO is 1:3, which means that 1 mole of Fe2O3 will react with 3 moles of CO.

Step 3: Calculate the number of moles of CO.
Number of moles of CO = Number of moles of Fe2O3 * stoichiometry ratio
Number of moles of CO = 2.845 mol * 3 = 8.535 mol

Step 4: Calculate the mass of CO.
Mass of CO = Number of moles of CO * molar mass of CO
Mass of CO = 8.535 mol * 28.01 g/mol = 238.53 g

Therefore, the mass of CO required to react with 454g of Fe2O3 is approximately 238.53 grams.

To determine the maximum amount of iron in grams that can be obtained from 454g (1.00 lb) of iron (III) oxide, we need to use stoichiometry.

Step 1: Convert the mass of Fe2O3 to moles.
The molar mass of Fe2O3 is calculated as:
(2 x molar mass of Fe) + (3 x molar mass of O)

The molar masses are:
Molar mass of Fe = 55.85 g/mol
Molar mass of O = 16.00 g/mol

Therefore, the molar mass of Fe2O3 is:
(2 x 55.85 g/mol) + (3 x 16.00 g/mol) = 159.70 g/mol

To convert the mass of Fe2O3 to moles, we divide the given mass by the molar mass:
454 g / 159.70 g/mol = 2.84 mol

Step 2: Use the stoichiometric coefficients to determine the moles of iron (Fe) produced.
From the balanced equation:
1 mole of Fe2O3 produces 2 moles of Fe

Thus, the number of moles of Fe produced will be twice the number of moles of Fe2O3:
2.84 mol Fe2O3 x (2 mol Fe / 1 mol Fe2O3) = 5.68 mol Fe

Step 3: Convert the moles of Fe to grams.
The molar mass of Fe is 55.85 g/mol.

To convert the moles of Fe to grams, we multiply the number of moles by the molar mass:
5.68 mol x 55.85 g/mol = 317.97 g

Therefore, the maximum mass of iron that can be obtained from 454 g of iron (III) oxide is 317.97 grams.

Now let's calculate the mass of CO required to react with the 454g of Fe2O3.

Step 1: Convert the mass of Fe2O3 to moles.
We already performed this calculation earlier, so we know that 454 g of Fe2O3 is equal to 2.84 mol.

Step 2: Use the stoichiometric coefficients to determine the moles of CO required.
From the balanced equation:
1 mole of Fe2O3 requires 3 moles of CO

So, to determine the moles of CO required, we multiply the moles of Fe2O3 by the appropriate stoichiometric ratio:
2.84 mol Fe2O3 x (3 mol CO / 1 mol Fe2O3) = 8.52 mol CO

Step 3: Convert the moles of CO to grams.
The molar mass of CO is calculated as:
(1 x molar mass of C) + (1 x molar mass of O)

The molar masses are:
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol

Thus, the molar mass of CO is:
(1 x 12.01 g/mol) + (1 x 16.00 g/mol) = 28.01 g/mol

To convert the moles of CO to grams, we multiply the number of moles by the molar mass:
8.52 mol x 28.01 g/mol = 238.81 g

Therefore, the mass of CO required to react with 454 g of Fe2O3 is 238.81 grams.