Question

Write balanced equations describing the reaction of sr with each of the following: ${\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{,}{\mathbf{\text{S}}}_{\mathbf{8}}\mathbf{,}{\mathbf{\text{Cl}}}_{\mathbf{2}}\mathbf{,}{\mathbf{\text{P}}}_{\mathbf{4}}\mathbf{,}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{,}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}$and $\mathit{H}\mathit{C}\mathit{I}$

Short answer

The balance equation reaction is

$$\begin{gathered} {\text{O}}_2:2{\text{Sr}}_{\left(v\right)}+{\text{O}}_{2\left(s\right)}\to 2{\text{Sr}}_{\left(v\right)} \\ {S}_8:8{\text{Sr}}_{\left(s\right)}+{\text{S}}_{\text{s}\left(\text{s}\right)}\to 8{\text{SrS}}_{\left(s\right)} \\ {\text{Cl}}_2:{\text{Sr}}_{\left(s\right)}+{\text{Cl}}_{2\left(s\right)}\to {\text{SrCl}}_{2\left(s\right)} \\ {P}_4:6{\text{Sr}}_{\left(s\right)}+{\text{P}}_{4\left(s\right)}\to 2{\text{Sr}}_3{\text{P}}_{2\left(s\right)} \\ {\text{H}}_2:{\text{Sr}}_{\left(s\right)}+{\text{H}}_{2\left(g\right)}\to {\text{SrH}}_{2\left(\theta \right)} \\ {\text{H}}_2\text{O}:{\text{Sr}}_{\left(s\right)}+{\text{H}}_2{\text{O}}_{\left(t\right)}\to {\text{E}}_{\left(aq\right)}^{2+}+2{\text{OH}}_{\left(\text{aqq}\right)}+{\text{H}}_{2\left(g\right)} \\ HCl:{\text{Sr}}_{\left(s\right)}+2{\text{HCl}}_{\left(\text{aq}\right)}\to {\text{SrCl}}_{2(\text{(aq)}}+{\text{H}}_{2\left(s\right)} \end{gathered}$$

Step-by-step solution

Definition of the equation reaction.

When the total charge and the number of atoms for each element in the reaction are the same for both the reactants and the products, the equation is said to be balanced. In other words, both sides of the reaction have the same mass and charge.

Determine the reactions.

The Group $2\text{A}$reduce ${\text{O}}_2$to form the oxides.

$2\text{Sr}\left(s\right)+{\text{O}}_2\left(g\right)\stackrel{}{\to }2\text{SrO}\left(s\right)$

The Group metals react with elemental sulfur to produce sulfides.

$8\text{Sr}\left(s\right)+{\text{S}}_{\text{g}}\left(s\right)\stackrel{}{\to }8\text{SrS}\left(s\right)$

The Group metals reduce halogens to form ionic halides.

$\text{Sr}\left(s\right)+{\text{Cl}}_2\left(\text{g}\right)\stackrel{}{\to }{\text{SrCl}}_2\left(s\right)$

Reduce the hydrogen:

Most of the Group $2\text{A}$metals reduce hydrogen to form an ionic hydride

$\text{Sr}\left(s\right)+{\text{H}}_2\left(g\right)\stackrel{}{\to }{\text{SrH}}_2\left(s\right)$

The larger Group $2\text{A}$metals reduce water to form hydrogen gas.

$\text{Sr}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\stackrel{}{\to }{\text{E}}^{2+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)+{\text{H}}_2\left(g\right)$

The Group $2\text{A}$metals react with ${\text{H}}^{+}$from the dissociation of $HCI$and produce hydrogen gas and the metal chloride.

$\text{Sr}\left(s\right)+2\text{HCl}\left(aq\right)\stackrel{}{\to }{\text{SrCl}}_2\left(aq\right)+{\text{H}}_2\left(g\right)$

Finally, the balance equation of $Sr$ .

$$\begin{gathered} {\text{O}}_2:2{\text{Sr}}_{\left(v\right)}+{\text{O}}_{2\left(s\right)}\to 2{\text{Sr}}_{\left(v\right)} \\ {S}_8:8{\text{Sr}}_{\left(s\right)}+{\text{S}}_{\text{s}\left(\text{s}\right)}\to 8{\text{SrS}}_{\left(s\right)} \\ {\text{Cl}}_2:{\text{Sr}}_{\left(s\right)}+{\text{Cl}}_{2\left(s\right)}\to {\text{SrCl}}_{2\left(s\right)} \\ {P}_4:6{\text{Sr}}_{\left(s\right)}+{\text{P}}_{4\left(s\right)}\to 2{\text{Sr}}_3{\text{P}}_{2\left(s\right)} \\ {\text{H}}_2:{\text{Sr}}_{\left(s\right)}+{\text{H}}_{2\left(g\right)}\to {\text{SrH}}_{2\left(\theta \right)} \\ {\text{H}}_2\text{O}:{\text{Sr}}_{\left(s\right)}+{\text{H}}_2{\text{O}}_{\left(t\right)}\to {\text{E}}_{\left(aq\right)}^{2+}+2{\text{OH}}_{\left(\text{aqq}\right)}+{\text{H}}_{2\left(g\right)} \\ HCl:{\text{Sr}}_{\left(s\right)}+2{\text{HCl}}_{\left(\text{aq}\right)}\to {\text{SrCl}}_{2(\text{(aq)}}+{\text{H}}_{2\left(s\right)} \end{gathered}$$