An illustration of two containers labelled as 'Vanadium (II) Oxide' and 'Iron (III) Oxide', next to them are two more containers labelled as 'Vanadium (V) Oxide' and 'Iron (II) Oxide'. The first two containers are tilted as if pouring their contents into a large chemical beaker in the middle. An action indicating a chemical reaction taking place is symbolized by bubbles and sparkles inside the beaker. Everything is set on a table with common laboratory equipment around, including test tubes, microscope, lab coats hanging nearby, and papers with chemistry notes scattered. No written text should be visible in the image.

Write the balanced equation for vandadium (II) oxide with iron (III) oxide results in vanadium (V) oxide and iron (II) oxide.

2VO + Fe2O3 - V2O5 + 6FeO2

Not sure if the is correct. We're having to learn this on our own with just what notes we have,

iron(II) oxide is FeO and the equation as you have written it is not balanced. (check the Fe--2 on left and 6 on right. O is not balanced either. 5 on left and 17 on right.

2VO + 3Fe2O3 -> V2O5 + 6FeO

Balanced equation is ✔️✔️✔️✔️

2VO + 3FeO3 -----> V2O5 + 6FeO

Well, your attempt at balancing the equation is not quite correct. Here's the corrected balanced equation:

2VO + 4Fe2O3 → V2O5 + 8FeO

Let's break it down to understand why:

To balance the equation, we need equal numbers of atoms on both sides.

On the left-hand side, we have:

- 2 vanadium (V) atoms (2VO)
- 8 oxygen atoms (4Fe2O3)
- 12 iron (III) atoms (4Fe2O3)

On the right-hand side, we have:

- 2 vanadium (V) atoms (V2O5)
- 5 oxygen atoms (V2O5)
- 8 iron (II) atoms (8FeO)

By adjusting the coefficients, we can balance the equation like this:

2VO + 4Fe2O3 → V2O5 + 8FeO

Now, we have:

- 2 vanadium (V) atoms on both sides
- 10 oxygen atoms on both sides
- 12 iron (III) atoms on the left and 8 iron (II) atoms on the right

Hope that clears things up for you! Let me know if you have any other questions.

To balance the equation correctly, we need to make sure that there is an equal number of atoms of each element on both sides of the equation.

Let's break down and balance the equation step by step:

1. Start by writing the chemical formulas of the reactants (vandadium (II) oxide and iron (III) oxide) and the products (vanadium (V) oxide and iron (II) oxide):
VO + Fe2O3 -> V2O5 + FeO

2. Count the number of atoms on both sides of the equation:
Reactants: V: 1, O: 3, Fe: 2
Products: V: 2, O: 5, Fe: 1

3. Balance the equation by adjusting the coefficients in front of each chemical formula. Start by balancing the elements that appear in the least amount first. In this case, we can start by balancing the vanadium atoms:
VO + Fe2O3 -> 2V2O5 + FeO

4. Now, let's balance the oxygen atoms. Since there are 3 oxygen atoms on the left side and 5 oxygen atoms on the right side, we can add a coefficient of 3 in front of the vanadium (II) oxide:
3VO + Fe2O3 -> 2V2O5 + FeO

5. Finally, balance the iron atoms. Since there are 2 iron atoms on the left side and 1 iron atom on the right side, we can add a coefficient of 2 in front of the iron (II) oxide:
3VO + Fe2O3 -> 2V2O5 + 2FeO

Therefore, the balanced equation for the reaction between vandadium (II) oxide and iron (III) oxide resulting in vanadium (V) oxide and iron (II) oxide is:
3VO + Fe2O3 -> 2V2O5 + 2FeO

I'm sorry, but that equation is not balanced. The correct balanced equation for the reaction between vanadium (II) oxide and iron (III) oxide to produce vanadium (V) oxide and iron (II) oxide is:

4 VO + 3 Fe2O3 -> 2 V2O5 + 6 FeO

This equation is balanced because there are the same number of atoms of each element on both sides of the reaction arrow.