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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.

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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.

From this link:

Reaction of europium with acids

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 $\ce{[Eu(OH2)9]^3+}$

$$\ce{2Eu(s) + 3H2SO4(aq) -> 2Eu^{3+}(aq) + 3SO4^{2-}(aq) + 3H2(g)}$$

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    $\begingroup$ With "excess sulfuric acid", sulfate ion does not exist either. Free acid will protonate sulfate ions to give bisulfate, which is close to but not exactly a strong acid. I would make bisulfate my anion and adjust coefficients accordingly. $\endgroup$ – Oscar Lanzi May 8 '16 at 21:43
  • $\begingroup$ @Oscar This is only true for highly concentrated solutions of sulfuric acid. It does not say which concentration was used, only that it was more than necessary to react europium fully. $\endgroup$ – Martin - マーチン Nov 8 '16 at 4:38
  • $\begingroup$ Sulfate/bisulfate buffers around pH 2, thus about 0.01 M solvated ptotons. At acid concentrations one would typically use bisulfate will predominate over free solvated protons. $\endgroup$ – Oscar Lanzi Nov 8 '16 at 9:35

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