# How to determine the equation for the reaction of europium with sulfuric acid based on experimental measurements?

I was given a set of results by a fellow researcher where he reacted $0.608~\mathrm{g}$ of europium ($M = 152$) with an excess of $\ce{H2SO4}$ and collected $144~\mathrm{cm^3}$ of $\ce{H2}$ gas at room temperature and pressure.

I was then asked to derive the equation for the reaction using these figures.

This is what I did.

I know that $1~\mathrm{mol}$ of gas is $24000~\mathrm{cm3}$. I found out that $144~\mathrm{cm^3}$ is equivalent to $0.006~\mathrm{mol}$ of $\ce{H2}$.

Then, I did $0.608 / 152$ to get $0.004~\mathrm{mol}$.

This meant that europium and hydrogen gas were in the ratio $2:3$.

The equation I wrote was $$\ce{2Eu + H2SO4 -> Eu2SO4 + 3H2}$$

But I am wrong and I don't know where.

Your equation isn't balanced; the oxidation state for europium is +3 and not +1; and europium sulfate does not exist in the solution you describe.

Europium metal dissolves readily in dilute sulphuric acid to form solutions containing the very pale pink aquated Eu(III) ion together with hydrogen gas, $$\ce{H2}$$. It is quite likely that $$\ce{Eu^3+}$$(aq) exists as largely the complex ion $$\ce{[Eu(OH2)9]^3+}$$
$$\ce{2Eu(s) + 3H2SO4(aq) -> 2Eu^{3+}(aq) + 3SO4^{2-}(aq) + 3H2(g)}$$