If 609.5 grams of vanadium (II) oxide, VO, and 832 grams of iron(III) oxide, Fe2O3, are put into a container and allowed to react according to the equation above, which substance(s) and how many grams of each would be present in the container after the reaction is complete?

To solve this question, we need to determine the balanced chemical equation for the reaction between vanadium (II) oxide (VO) and iron (III) oxide (Fe2O3). The equation should give us information about the stoichiometry of the reaction, which will allow us to calculate the quantities of substances present after the reaction.

The balanced equation for the reaction between VO and Fe2O3 is:

3VO + Fe2O3 -> 3V2O3 + 2Fe

According to the equation, when 3 moles of VO react with 1 mole of Fe2O3, they produce 3 moles of V2O3 and 2 moles of Fe.

Now, let's calculate the molar masses of VO and Fe2O3:

- Molar mass of VO = (atomic mass of V) + (atomic mass of O) = (50.9 g/mol) + (16.0 g/mol) = 66.9 g/mol
- Molar mass of Fe2O3 = (2 * atomic mass of Fe) + (3 * atomic mass of O) = (2 * 55.8 g/mol) + (3 * 16.0 g/mol) = 159.7 g/mol

Now we can use the molar masses and the given masses to calculate the number of moles of VO and Fe2O3:

- Moles of VO = mass of VO / molar mass of VO = 609.5 g / 66.9 g/mol ≈ 9.11 mol
- Moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3 = 832 g / 159.7 g/mol ≈ 5.21 mol

Now, we can determine the limiting reactant, which is the reactant that is completely consumed in the reaction. The reactant that produces fewer moles of product is the limiting reactant.

From the balanced equation, it is clear that 3 moles of VO react with 1 mole of Fe2O3 to produce 3 moles of V2O3. Therefore, for every 1 mole of Fe2O3, we will need 3/1 moles of VO.

Since we have 5.21 moles of Fe2O3, the corresponding moles of VO needed would be (3/1) * 5.21 ≈ 15.63 moles of VO.

However, we only have 9.11 moles of VO available, which is less than what is needed. This means VO is the limiting reactant, and Fe2O3 is in excess.

To calculate the masses of substances present after the reaction, we need to use the stoichiometry of the reaction.

From the balanced equation, we can see that 3 moles of VO react to form 3 moles of V2O3. Therefore, the number of moles of V2O3 produced will be equal to the number of moles of VO used, which is 9.11 mol.

To find the mass of V2O3, we can use its molar mass:

- Molar mass of V2O3 = (2 * atomic mass of V) + (3 * atomic mass of O) = (2 * 50.9 g/mol) + (3 * 16.0 g/mol) = 149.9 g/mol

Therefore, the mass of V2O3 produced will be:

Mass of V2O3 = moles of V2O3 * molar mass of V2O3 = 9.11 mol * 149.9 g/mol ≈ 1365 g

Since Fe2O3 is in excess, we can assume that it is not fully consumed. Therefore, the total mass of Fe2O3 will remain the same, which is 832 grams.

In summary, after the reaction is complete, there will be approximately 1365 grams of V2O3 and 832 grams of Fe2O3 remaining in the container.

To determine the substances and their respective amounts present after the reaction is complete, we need to balance the chemical equation first:

2 Fe2O3 + 5 VO → 4 FeO + V2O5

The balanced equation shows that 2 moles of Fe2O3 react with 5 moles of VO to produce 4 moles of FeO and 1 mole of V2O5.

Now, let's calculate the moles of each substance in the given samples:

For VO:
- Mass of VO = 609.5 grams
- Molar mass of VO = 50.94 g/mol (atomic mass of V: 50.94 g/mol, atomic mass of O: 16.00 g/mol)
- Moles of VO = Mass of VO / Molar mass of VO = 609.5 g / 50.94 g/mol ≈ 11.97 mol

For Fe2O3:
- Mass of Fe2O3 = 832 grams
- Molar mass of Fe2O3 = 159.69 g/mol (atomic mass of Fe: 55.85 g/mol, atomic mass of O: 16.00 g/mol)
- Moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3 = 832 g / 159.69 g/mol ≈ 5.20 mol

According to the balanced equation, the reactants will be consumed in the stoichiometric ratio of 5 moles of VO to 2 moles of Fe2O3. However, we have an excess of VO:

Excess reacting material:
- Moles of VO remaining = Moles of VO - (Moles of Fe2O3 × Coefficient ratio) = 11.97 mol - (5.20 mol × 5/2) ≈ 1.47 mol

Products formed:
- Moles of FeO formed = (Moles of Fe2O3 × Coefficient ratio) = 5.20 mol × 4/2 = 10.40 mol
- Moles of V2O5 formed = (Moles of Fe2O3 × Coefficient ratio) / 2 = 5.20 mol × 1/2 = 2.60 mol

To calculate the mass of each substance, multiply the number of moles by their respective molar masses:

For VO remaining:
- Mass of VO remaining = Moles of VO remaining × Molar mass of VO = 1.47 mol × 50.94 g/mol ≈ 74.94 grams

For FeO formed:
- Mass of FeO formed = Moles of FeO formed × Molar mass of FeO = 10.40 mol × 71.85 g/mol ≈ 746.04 grams

For V2O5 formed:
- Mass of V2O5 formed = Moles of V2O5 formed × Molar mass of V2O5 = 2.60 mol × 181.88 g/mol ≈ 473.49 grams

Therefore, after the reaction is complete, there would be approximately 74.94 grams of VO remaining, 746.04 grams of FeO formed, and 473.49 grams of V2O5 formed in the container.