Consider the reaction: 2AI(s) + 6H+ (aq) → 2AI3+ (aq) + 3H2(g). What is the total number of moles of electrons transferred from the aluminium atoms to the hydrogen ions?
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For the reaction $2\text{Al}(s) + 6\text{H}^+ (aq) \rightarrow 2\text{Al}^{3+} (aq) + 3\text{H}_2(g)$:
Look at what each Al atom does:
$\text{Al} \rightarrow \text{Al}^{3+} + 3e^-$
Each aluminium atom loses 3 electrons to form $\text{Al}^{3+}$.
You have 2 Al atoms in the equation, so:
$2 \text{ atoms} \times 3 e^- \text{ per atom} = 6 \text{ moles of electrons}$ total.
Check with hydrogen: $6\text{H}^+ + 6e^- \rightarrow 3\text{H}_2$. That matches: 6 electrons gained by H⁺.
So 6 moles of electrons are transferred from aluminium to the hydrogen ions for the reaction as written.
For the reaction 2AI(s) + 6H⁺ (aq) → 2AI³⁺ (aq) + 3H2(g)
Look at what each Al atom does:
Al ------ Al³⁺ + 3e‐
Each aluminium atom loses 3 electrons to form Al³⁺.
You have 2 Al atoms in the equation, so:
2 atoms × 3e‐ per atom = 6 moles of electrons total.
Check with Hydrogen: 6H⁺ + 6e‐ = 3H2.
That matches: 6 electrons gained by H⁺
So 6 moles of electrons are transferred from aluminium to the hydrogen ions for the reaction as written.
Translated by Bassey mathJx.
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