# Stoichiometric coefficients in reaction between vanadium(II) oxide and iron(III) oxide

Calculate the weight of $$\ce{V2O5}$$ produced from $$\pu{2g}$$ of $$\ce{VO}$$ and $$\pu{5.75 g}$$ of $$\ce{Fe2O3}$$

Now we write the reaction as following - $$\ce{2 VO + 3Fe2O3-> V2O5 + 6FeO}$$

Now, by the Principle of Atom Conservation on vanadium, we get,

$$n_\ce{VO} = 2\times n_\ce{V2O5}$$ $$\implies \pu{2 g} / \pu{67 g mol-1} = 2\times n_\ce{V2O5}$$ $$\implies n_\ce{V2O5} = \frac{\pu{0.029850 mol}}{2} = \pu{0.014925 mol}$$ $$m_\ce{V2O5} = \pu{0.0149250 mol} * \pu{182 g mol-1} \approx \pu{2.71 g}$$

I know that this can be done by limiting reagent, but what is wrong in this? The correct answer is $$\boxed{\pu{2.18 g}}$$

• Your calculation is correct if all $\ce{VO}$ is transformed into $\ce{V2O5}$. But here, the amount of $\ce{Fe2O3}$ is not sufficient to transform all $\ce{VO}$ into $\ce{V2O5}$. So please start your calculation again starting with the amount of $\ce{Fe2O3}$ Aug 25, 2022 at 20:48
• @Karsten, the edit that balanced the reaction given by the OP deviates from the original intent. They did not actually plan to balance and solve. Aug 26, 2022 at 6:24
• @Aqua Note that your task analysis has been very rushed. There is a saying "Measure twice, cut once". Applied to practical tasks - "Think twice, calculate once." Aug 26, 2022 at 7:51

Within a chemical reaction, no new atoms are created or destroyed. More formally, $$\sum{m_\mathrm{before\,rxn} = \sum{m_\mathrm{after\, rxn}}}$$

Moving onto what you have done

Now, by the Principle of Atom Conservation on vanadium, we get,

$$n_\ce{VO} = 2 \times n_\ce{V2O5}$$ $$\implies 2/67 = 2\times n_\ce{V2O5}$$

This is correct (well kind of correct). However, this makes one assumption that doesn't really exist in this scenario. It assumes that all the $$\ce{VO}$$ reacts within the reaction. Which unfortunately doesn't happen here.

Taken from Rafael L's comment, the right method to do this (and the most accurate method) is as follows

[...] convert the masses of $$\ce{VO}$$ and $$\ce{Fe2O3}$$ to moles, figure out which is the limiting reagent (balanced eqn: $$\ce{2VO + 3Fe2O3 → 6FeO + V2O5}$$), then calculate the moles of $$\ce{V2O5}$$ produced and convert moles of $$\ce{V2O5}$$ to mass.

Now, following this, we see the number of equivalents on $$\ce{VO}$$ and $$\ce{Fe2O3}$$ are as follows,

$$n_\ce{Fe2O3} = \pu{0.03593 mol} \implies \mathrm{equiv}_\ce{Fe2O3} = \pu{0.011979 eq}$$ $$n_\ce{VO} = \pu{0.029850 mol} \implies \mathrm{equiv}_\ce{VO} = \pu{0.0149250 eq}$$

From here, you can notice that the amount of $$\ce{Fe2O3}$$ that reacts is more than the $$\ce{VO}$$ but the number of equivalents is less. What this implies is that there is some $$\ce{VO}$$ left over. The rest is just multiplication.

[...] doesn't then POAC fail entirely? I used it many times which involved such reactions.

The answer to this question is no. POAC still holds. It's just a bit more complicated. Let's check this with the formal definition,

$$\sum{m_\mathrm{before \, rxn} = \sum{m_\mathrm{after \, rxn}}}$$ \begin{align} m_\ce{VO} &= \pu{0.395 g}\\ m_\ce{Fe2O3} &= \pu{0 g}\\ m_\ce{V2O5} &= \pu{2.180 g}\\ m_\ce{FeO} &= \pu{5.174 g}\\ \end{align}

\begin{align} m_\mathrm{before\,rxn} &= \pu{7.75 g} \\ m_\mathrm{after\,rxn} &= \pu{7.749 g} \end{align}

Therefore, both are approximately equal, and POAC holds.

[OP] Now, by the Principle of Atom Conservation on vanadium, we get, $$n_\ce{VO} = 2\times n_\ce{V2O5}$$

This is not universally true. In the extreme case where you don't provide any iron, it should be easy to see: There is no reaction, so none of the product is made even though there is reactant. Also, the notation is a bit sloppy. You have to know that the first term is supposed to be about "before the reaction", and the second term about "after the reaction". Some textbooks use ICE tables to force you to make this explicit, by writing the amounts in the appropriate box.

The clearer way to write this is as follow:

$$- \Delta n_\ce{VO} = 2\times \Delta n_\ce{V2O5}$$

In words, for each mole of $$\ce{V2O5}$$ produced, 2 moles of $$\ce{VO}$$ are consumed in this reaction. Or, and this is easier for me personally because it visually corresponds to the statement just made, you can set up the ratio:

$$\frac{\Delta n_\ce{VO}}{\Delta n_\ce{V2O5}} = \frac{\pu{-2 mol}}{\pu{1 mol}}$$

It is possible that the reaction does not happen at all or does not use up all the $$\ce{VO}$$, then this relationship still holds. However, if there is a different reaction (or a side reaction), you can no longer rely on this modified relationship either.

[OP] I know that this can be done by limiting reagent, but what is wrong in this?

If your reactant is limiting, you can set $$\Delta n_\ce{VO}$$ to $$-n_{0,\ce{VO}}$$, i.e. all gets used up when the reaction goes to completion. If your reactant is in excess, you have to figure out how much reacts, i.e. what $$\Delta n_\ce{VO}$$ is. In either case, no atoms are created or destroyed, though.