Question

1. the reaction between solid aluminum and liquid bromine is exothermic. this reaction creates the solid product aluminum bromide. what equation includes the correct formulas and demonstrates conservation of mass? \\( 2\text{al}(s) + 3\text{br}_2(l) \

ightarrow 2\text{albr}_3(s) \\) \\( \text{al}(s) + \text{br}_2(l) \
ightarrow \text{albr}_3(s) \\) \\( \text{al}(s) + \text{br}(l) \
ightarrow \text{albr}(s) \\) \\( \text{al}(s) + 3\text{br}(l) \
ightarrow \text{albr}_3(s) \\)

Explanation:

Step1: Identify correct formulas

Aluminum is $\text{Al}(s)$, bromine exists as diatomic liquid $\text{Br}_2(l)$, aluminum bromide is $\text{AlBr}_3(s)$ (Al has +3 charge, Br has -1 charge, so 3 Br⁻ balance 1 Al³⁺). This eliminates options C and D (wrong bromine formula).

Step2: Balance the equation

Count atoms on each side:

• Unbalanced: $\text{Al}(s) + \text{Br}_2(l)

ightarrow \text{AlBr}_3(s)$

• Balance Br: 3 $\text{Br}_2$ gives 6 Br, so need 2 $\text{AlBr}_3$ (6 Br total).
• Balance Al: 2 $\text{Al}$ to match 2 Al in $2\text{AlBr}_3$.

Final balanced equation: $2\text{Al}(s) + 3\text{Br}_2(l)
ightarrow 2\text{AlBr}_3(s)$, which matches option A.

Answer:

A. $2\text{Al}(s) + 3\text{Br}_2(l)
ightarrow 2\text{AlBr}_3(s)$