Over the years, the thermite reaction (mixing of solid iron (III) oxide with aluminum metal) has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors.

A. Write a balanced equation representing the reaction.

B. What masses of iron (III) oxide and aluminum must be used to produce 15.0 g if iron?

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

Complete and balance this reaction to form vanadium (V) oxide

V(s) + O2(g) �

@DrBob222 if you could either change your response or let someone else respond that would be great because either you just are incapable of explaining things in a way that would make sense or you are wrong.

A. To write a balanced equation representing the reaction, we need to determine the chemical formulas and the stoichiometric coefficients of the reactants and products.

The reactants are solid iron (III) oxide (Fe2O3) and aluminum metal (Al). The products are solid iron (Fe) and aluminum oxide (Al2O3).

The balanced equation for the reaction can be written as:

Fe2O3 + 2Al -> 2Fe + Al2O3

B. To determine the masses of iron (III) oxide and aluminum needed to produce 15.0 g of iron, we can use the stoichiometry of the balanced equation.

Let's assume that x grams of Fe2O3 and y grams of Al are required.

From the equation, we can see that the molar ratio between Fe2O3 and Fe is 1:2. This means that 1 mole of Fe2O3 reacts with 2 moles of Fe.

The molar mass of Fe2O3 is (2 × atomic mass of Fe) + (3 × atomic mass of O) = (2 × 55.845 g/mol) + (3 × 16.00 g/mol) = 159.69 g/mol.

Similarly, the molar mass of Fe is 55.845 g/mol.

Using the molar ratio and the molar masses, we can set up the following equation:

(x g Fe2O3) / (159.69 g/mol) = (15.0 g Fe) / (55.845 g/mol)

Solving for x, we find that x ≈ 42.16 g.

Next, we can use the molar ratio between Al and Fe2O3, which is 1:2. This means that 2 moles of Al react with 1 mole of Fe2O3.

The molar mass of Al is 26.98 g/mol.

Using the molar ratio and the molar mass, we can set up the following equation:

(y g Al) / (26.98 g/mol) = (42.16 g Fe2O3) / (159.69 g/mol)

Solving for y, we find that y ≈ 33.29 g.

Therefore, to produce 15.0 g of iron, approximately 42.16 g of iron (III) oxide and 33.29 g of aluminum should be used.

C. To calculate the maximum mass of aluminum oxide that could be produced, we need to use the same stoichiometry as in part B.

From the balanced equation, we can see that the molar ratio between Al and Al2O3 is 4:2, or 2:1. This means that 2 moles of Al produce 1 mole of Al2O3.

The molar mass of Al2O3 is (2 × atomic mass of Al) + (3 × atomic mass of O) = (2 × 26.98 g/mol) + (3 × 16.00 g/mol) = 101.96 g/mol.

Using the molar ratio and the molar mass, we can set up the following equation:

(y g Al) / (26.98 g/mol) = (m g Al2O3) / (101.96 g/mol)

Solving for m, we find that m = (2/3) × y ≈ (2/3) × 33.29 g ≈ 22.19 g.

Therefore, the maximum mass of aluminum oxide that could be produced is approximately 22.19 g.

Fe2O3 + 2Al ==> Al2O3 + 2Fe

B.
a. mols Fe in 15.0g Fe = grams/molar mass
b. Using the coefficients in the balanced equation, convert mols Fe to mols Fe2O3.
c. Convert mols Fe2O3 to g. g = mols x molar mass

For Al, use the same procedure but in b convert mols Fe to mols Al and go from therre.

C.
Same procedure. Convert mols Fe to mols Al2O3 then convert to grams Al2O3.