# Question a5e9d

Jul 2, 2016

Here's what I got.

#### Explanation:

Calcium metal, $\text{Ca}$, will react with bromine, ${\text{Br}}_{2}$, to form calcium bromide, ${\text{CaBr}}_{2}$, a white solid.

${\text{Ca"_ ((s)) + "Br"_ (2(l)) -> "CaBr}}_{2 \left(s\right)}$

This is a redox reaction in which calcium is being oxidized, its oxidation state going from $\textcolor{b l u e}{0}$ to $\textcolor{b l u e}{+ 2}$, and bromine is being reduced, its oxidation state going from $\textcolor{b l u e}{0}$ to $\textcolor{b l u e}{- 1}$.

${\stackrel{\textcolor{b l u e}{0}}{\text{Ca")_ ((s)) + stackrel(color(blue)(0))("Br") _ (2(l)) -> stackrel(color(blue)(+2))("Ca") stackrel(color(blue)(-1))("Br}}}_{2 \left(s\right)}$

Now, I think that this reaction actually involves gaseous bromine. Calcium metal burns in the presence of bromine gas to form calcium bromide

${\text{Ca"_ ((s)) + "Br"_ (2(g)) -> "CaBr}}_{2 \left(s\right)}$

As a final note, the aqueous state suggests that the compound is dissolved in water. Therefore, you cannot use the aqueous state unless water is present.

For example, another reaction used to synthesize calcium bromide involves the neutralization of calcium hydroxide, "Ca"("OH")_2#, with hydrobromic acid, $\text{HBr}$

${\text{Ca"("OH")_ (2(s)) + 2"HBr"_ ((aq)) -> "CaBr"_ (2(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

This time, the reaction produces aqueous calcium bromide because the product is dissolved in water, i.e. it exists in aqueous solution as calcium cations and bromide anions

${\text{CaBr"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + 2"Br}}_{\left(a q\right)}^{-}$