Equation: Fe2O3 + 2Al ---> 2Fe + Al2O3.
How many grams of Fe2O3 react with excess Al to make 475g Fe?
Can you show me step by step how to do it? My book says the answer is 679g Fe2O3.
The book is right.
ALL of these problems can be worked this way.
1. Write and balance the equation.
2. Convert what you have (in this case Fe) to moles. moles = grams/molar mass= ??.
3. Using the coefficients in the balanced equation, convert moles of what you have into moles of what you want (in this case Fe2O3).
moles Fe2O3 = moles Fe x (1 mole Fe2O3/2 moles Fe) = moles Fe x 1/2 = ??
4. Convert moles of what you have to the units you want (in this case grams). grams = moles x molar mass.
I get 679.0756 g which rounds to 679 to three significant figures (the number 675 limits you to 3 s.f.).
Sure, let's solve this step-by-step!
First, we need to determine the molar mass of Fe2O3 and Fe because we'll be working with moles.
The molar mass of Fe2O3 is:
(2 atoms of Fe × atomic mass of Fe) + (3 atoms of O × atomic mass of O)
= (2 × 55.847 g/mol) + (3 × 16.00 g/mol)
= 111.694 g/mol + 48.00 g/mol
= 159.694 g/mol
The molar mass of Fe is:
1 atom of Fe × atomic mass of Fe
= 1 × 55.847 g/mol
= 55.847 g/mol
Next, we can calculate the number of moles of Fe we want to produce:
mols of Fe = mass of Fe / molar mass of Fe
mols of Fe = 475 g / 55.847 g/mol
mols of Fe ≈ 8.50 mol Fe
Now, let's use the balanced equation to determine the molar ratio between Fe2O3 and Fe.
According to the balanced equation, the ratio of Fe2O3 to Fe is 1:2. This means that for every 1 mole of Fe2O3, we get 2 moles of Fe.
Therefore, we can set up a proportion to figure out how many moles of Fe2O3 are needed:
1 mol Fe2O3 / 2 mol Fe = x mol Fe2O3 / 8.50 mol Fe
Cross-multiplying and solving for x, we get:
x mol Fe2O3 = (1 mol Fe2O3 / 2 mol Fe) × 8.50 mol Fe
x mol Fe2O3 = 4.25 mol Fe2O3
Finally, we can convert the moles of Fe2O3 to grams:
mass of Fe2O3 = molar mass of Fe2O3 × moles of Fe2O3
mass of Fe2O3 = 159.694 g/mol × 4.25 mol
mass of Fe2O3 ≈ 679 g Fe2O3
So, according to the calculations, approximately 679 grams of Fe2O3 will react with an excess of Al to produce 475 grams of Fe.
To find the number of grams of Fe2O3 that react with excess Al to produce 475g Fe, you can use the balanced chemical equation and the concept of stoichiometry.
Step 1: Write down the balanced chemical equation:
Fe2O3 + 2Al ---> 2Fe + Al2O3
Step 2: Determine the molar mass of Fe2O3:
Fe2O3 = (2 x atomic mass of Fe) + (3 x atomic mass of O)
Fe2O3 = (2 x 55.845 g/mol) + (3 x 16.00 g/mol)
Fe2O3 = 159.69 g/mol
Step 3: Use the stoichiometry of the reaction to convert grams of Fe to grams of Fe2O3.
Since the stoichiometric ratio is 2:1 (2 moles of Fe2O3 react with 1 mole of Fe), you can set up a proportion to solve for the unknown amount of Fe2O3, x.
(475 g Fe / 2 Fe) = (x g Fe2O3 / 1 Fe2O3)
Step 4: Simplify and solve the proportion:
(475 g Fe / 2 Fe) = x g Fe2O3
x = (475 g Fe / 2 Fe) * (1 Fe2O3 / 1 g Fe2O3)
x = 237.5 g Fe2O3
So, the correct number of grams of Fe2O3 required to react with excess Al to produce 475g Fe is 237.5 g Fe2O3, which is not the same as the answer given in your book (679g). It is possible that there is an error in the book or in the problem statement.
To determine the amount of Fe2O3 required to produce 475g of Fe, we can use stoichiometry. Stoichiometry allows us to relate the amounts of reactants and products in a chemical equation.
Let's break down the problem step by step:
Step 1: Write down the balanced chemical equation.
The balanced equation given is:
Fe2O3 + 2Al ---> 2Fe + Al2O3
This equation shows that 1 mole of Fe2O3 reacts with 2 moles of Al to produce 2 moles of Fe and 1 mole of Al2O3.
Step 2: Calculate the molar mass of Fe.
The molar mass of Fe is 55.85 g/mol. Since we are trying to find the mass of Fe, it is essential to know the molar mass for accurate calculations.
Step 3: Calculate the number of moles of Fe.
We are given the mass of Fe, which is 475g. To convert grams to moles, use the equation:
Number of moles = mass / molar mass
Number of moles of Fe = 475g / 55.85 g/mol ≈ 8.50 mol
Step 4: Determine the stoichiometric ratio.
From the balanced equation, we can see that:
1 mole of Fe2O3 reacts with 2 moles of Fe
Therefore, the number of moles of Fe2O3 required can be calculated using the ratio:
Number of moles of Fe2O3 = (Number of moles of Fe) / 2
Number of moles of Fe2O3 = 8.50 mol / 2 ≈ 4.25 mol
Step 5: Convert moles to grams.
To convert moles to grams, use the equation:
Mass = Number of moles * Molar mass
Mass of Fe2O3 = 4.25 mol * (2 * (55.85 g/mol) + 3 * (16.00 g/mol))
Mass of Fe2O3 ≈ 679g
Hence, the answer is indeed 679g Fe2O3.