Question

question 5 (1 point)
in the reaction: $ce{al}(s) + ce{o_2}(g) \
ightarrow ce{al_2o_3}(s)$, how many electrons does an atom of aluminum lose or gain?
\bigcirc aluminum loses 1 electron
\bigcirc aluminum gains 1 electron
\bigcirc aluminum loses 3 electrons
\bigcirc aluminum gains 3 electrons
\bigcirc there is no loss or gain of electrons

Explanation:

In the reaction \( \text{Al}(s) + \text{O}_2(g)
ightarrow \text{Al}_2\text{O}_3(s) \), aluminum (Al) is a metal. In ionic compounds like \( \text{Al}_2\text{O}_3 \), aluminum typically forms a \( +3 \) oxidation state. To achieve this, an aluminum atom loses 3 electrons (since losing electrons increases the oxidation state; the oxidation state of Al in the reactant is 0, and in the product, it is \( +3 \), so the change is \( +3 \), meaning it loses 3 electrons).

Answer:

C. aluminum loses 3 electrons (assuming the options are labeled as A to E with C being "aluminum loses 3 electrons")