Given the reaction:

Fe2O3 + Al -> Fe Al2O3

What masses of iron oxide and aluminum must be used to produce 15g iron?

What is the maximum mass of aluminum oxide that could be produced?

How much aluminum oxide would be produced if the yield is 93%?

Please show solutions. :]

One has to start by balancinge the equation. Do that.

Then convert the 15g of iron to moles. Do that. THen, you know the mole relations ships between the iron and the reactants, and how many moles of reactants you need.

We will be happy to critique your work.

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The balanced equation would be

Fe2O3 Al -> Fe + Al2O3
15g of iron = .03 moles (rounded-off)

then what?

To answer these questions, we need to use stoichiometry and molar mass calculations. Let's start by calculating the molar masses:

Molar mass of Fe2O3:
Fe: 2 * atomic mass of Fe = 2 * 55.845 g/mol = 111.69 g/mol
O: 3 * atomic mass of O = 3 * 15.999 g/mol = 47.997 g/mol

Molar mass of Al:
Al: atomic mass of Al = 26.982 g/mol

Molar mass of Al2O3:
Al: 2 * atomic mass of Al = 2 * 26.982 g/mol = 53.964 g/mol
O: 3 * atomic mass of O = 3 * 15.999 g/mol = 47.997 g/mol

Now let's calculate the quantities:

Question 1: What masses of iron oxide and aluminum must be used to produce 15g iron?

Since the molar ratio of Fe2O3 to Fe is 1:2, we can set up the following equation based on the balanced equation:

(2 mol Fe / 1 mol Fe2O3) * (molar mass of Fe / molar mass of Fe2O3) * (mass of Fe2O3) = mass of Fe

Let's substitute the known values into the equation:

(2 mol Fe / 1 mol Fe2O3) * (55.845 g/mol / 159.69 g/mol) * (mass of Fe2O3) = 15 g

Simplifying the equation:

(2) * (0.3495) * (mass of Fe2O3) = 15
0.699 * mass of Fe2O3 = 15
mass of Fe2O3 = 15 / 0.699 = 21.516 g

So, you would need approximately 21.516 grams of iron oxide (Fe2O3) to produce 15 grams of iron.

Question 2: What is the maximum mass of aluminum oxide that could be produced?

Since the molar ratio of Al2O3 to Al is 1:2, we can set up the following equation based on the balanced equation:

(2 mol Al2O3 / 2 mol Al) * (molar mass of Al2O3 / molar mass of Al) * (mass of Al) = mass of Al2O3

Let's substitute the known values into the equation:

(2 mol Al2O3 / 2 mol Al) * (53.964 g/mol / 26.982 g/mol) * (mass of Al) = mass of Al2O3

Simplifying the equation:

(1) * (1.996) * (mass of Al) = mass of Al2O3

So, the maximum mass of aluminum oxide (Al2O3) that could be produced is approximately 1.996 times the mass of aluminum (Al).

Question 3: How much aluminum oxide would be produced if the yield is 93%?

If the yield is 93%, it means that 93% of the theoretical maximum mass of aluminum oxide will be obtained.

To calculate the mass of aluminum oxide produced, we can simply multiply the maximum mass of aluminum oxide by the yield percentage:

mass of Al2O3 = (93% / 100%) * (maximum mass of Al2O3)

Substitute the known values into the equation:

mass of Al2O3 = (0.93) * (maximum mass of Al2O3)

So, the amount of aluminum oxide produced, with a 93% yield, will be 93% of the maximum mass of aluminum oxide.

To answer these questions, we need to use stoichiometry and molar mass calculations.

First, let's determine the molar masses of the compounds involved:

- Molar mass of Fe2O3 (iron oxide) = atomic mass of iron (Fe) + (3 * atomic mass of oxygen (O))
= (55.845 g/mol) + (3 * 16.00 g/mol)
= 159.69 g/mol

- Molar mass of Al (aluminum) = 26.98 g/mol

Now, let's address each question:

1. What masses of iron oxide and aluminum must be used to produce 15g iron?

From the balanced equation, we know that the ratio of the coefficients is 1:1. This means that for every 1 mole of iron produced, we need 1 mole of iron oxide and 1 mole of aluminum.

To calculate the moles of iron, we convert the given mass to moles:
Moles of Fe = Mass of Fe / Molar mass of Fe
= 15g / 55.845 g/mol
≈ 0.268 mol

Since the ratio is 1:1, the moles of iron oxide and aluminum are also 0.268 mol.

To calculate the masses, we multiply the moles by their respective molar masses:
Mass of iron oxide = Moles of Fe2O3 * Molar mass of Fe2O3
= 0.268 mol * 159.69 g/mol
≈ 42.8 g

Mass of aluminum = Moles of Al * Molar mass of Al
= 0.268 mol * 26.98 g/mol
≈ 7.24 g

2. What is the maximum mass of aluminum oxide that could be produced?

Since the ratio from the balanced equation is 1:1, the moles of aluminum oxide produced will be the same as the moles of iron produced, which we calculated as 0.268 mol.

To determine the mass, we multiply the moles of aluminum oxide by its molar mass:
Mass of aluminum oxide = Moles of Al2O3 * Molar mass of Al2O3
= 0.268 mol * (2 * 26.98 g/mol + 3 * 16.00 g/mol)
≈ 79.89 g

3. How much aluminum oxide would be produced if the yield is 93%?

The yield represents the percentage of the theoretical maximum amount of product that is obtained in an actual experiment. In this case, the yield is given as 93%.

We can calculate the actual mass of aluminum oxide produced by multiplying the theoretical maximum mass (79.89 g) by the yield percentage:
Actual mass of aluminum oxide = 79.89 g * 0.93
≈ 74.31 g

Therefore, if the yield is 93%, approximately 74.31 g of aluminum oxide would be produced.

I hope this helps! Let me know if you have any further questions.