I have a worksheet that has a question on it the question states: How much iron can be recovered from 25.0 g of Fe2O3 and that is all i have i have just started in the middle of this class can i have an explanation of how to get to the answer please?
mols Fe2O3 = grams/molar mass = ?
You can get 2 mol Fe for every mole of Fe2O3. So 2 x ? = mols Fe.
Certainly! To determine how much iron can be recovered from 25.0 g of Fe2O3, we need to use the concept of stoichiometry.
Stoichiometry is a way to calculate the quantities of reactants and products in a chemical reaction. In this case, we are trying to find the amount of iron (Fe) that can be obtained from a given amount of iron(III) oxide (Fe2O3).
Here's the step-by-step process:
Step 1: Write the balanced chemical equation for the reaction.
The balanced equation for the reaction between iron(III) oxide and elemental iron can be represented as:
2 Fe2O3 + 3 C → 4 Fe + 3 CO2
This equation tells us that for every 2 moles of Fe2O3, we can obtain 4 moles of Fe.
Step 2: Calculate the molar mass of Fe2O3 and Fe.
The molar mass of a substance is the mass of one mole of that substance. The molar mass of Fe2O3 is approximately 159.7 g/mol, and the molar mass of Fe is 55.8 g/mol.
Step 3: Convert the given mass of Fe2O3 to moles.
To do this, divide the mass of Fe2O3 (25.0 g) by its molar mass (159.7 g/mol):
25.0 g / 159.7 g/mol = 0.1567 mol
Step 4: Determine the moles of Fe using the mole ratio from the balanced equation.
Since the mole ratio between Fe2O3 and Fe is 2:4 (from the balanced equation), we can use this ratio to find the moles of Fe.
0.1567 mol (Fe2O3) x (4 mol Fe / 2 mol Fe2O3) = 0.3134 mol Fe
Step 5: Calculate the mass of Fe obtained.
To determine the mass of Fe, multiply the moles of Fe (0.3134 mol) by its molar mass (55.8 g/mol):
0.3134 mol x 55.8 g/mol = 17.5 g
Therefore, the amount of iron that can be recovered from 25.0 g of Fe2O3 is approximately 17.5 g.
It's worth mentioning that this calculation assumes 100% yield, meaning that all the Fe2O3 is completely converted to Fe. In practice, the actual yield may be lower due to side reactions or incomplete conversion.