Write balanced equations describing the reaction of Sr with each of the following: \(\mathrm{O}_{2}, \mathrm{S}, \mathrm{Cl}_{2}, \mathrm{P}_{4}, \mathrm{H}_{2}, \mathrm{H}_{2} \mathrm{O},\) and HCl.

Short Answer

Expert verified
Here are the balanced chemical equations for the reactions of Strontium (Sr) with various substances: 1. With Oxygen (O₂): \[ 2 \mathrm{Sr} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{SrO} \] 2. With Sulfur (S): \[ \mathrm{Sr} + \mathrm{S} \longrightarrow \mathrm{SrS} \] 3. With Chlorine (Cl₂): \[ \mathrm{Sr} + \mathrm{Cl}_{2} \longrightarrow \mathrm{SrCl}_{2} \] 4. With Phosphorus (P₄): \[ 6 \mathrm{Sr} + 2 \mathrm{P}_{4} \longrightarrow 3 \mathrm{Sr}_{3}\mathrm{P}_{2} \] 5. With Hydrogen (H₂): \[ \mathrm{Sr} + \mathrm{H}_{2} \longrightarrow \mathrm{SrH}_{2} \] 6. With Water (H₂O): \[ \mathrm{Sr} + 2 \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{Sr}(\mathrm{OH})_{2} + \mathrm{H}_{2} \] 7. With Hydrochloric acid (HCl): \[ \mathrm{Sr} + 2 \mathrm{HCl} \longrightarrow \mathrm{SrCl}_{2} + \mathrm{H}_{2} \]

Step by step solution

01

Reaction with Oxygen (O₂)

When Strontium reacts with Oxygen, it forms Strontium Oxide (SrO). The balanced equation will be: \[ 2 \mathrm{Sr} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{SrO} \]
02

Reaction with Sulfur (S)

When Strontium reacts with Sulfur, it forms Strontium Sulfide (SrS). The balanced equation will be: \[ \mathrm{Sr} + \mathrm{S} \longrightarrow \mathrm{SrS} \]
03

Reaction with Chlorine (Cl₂)

When Strontium reacts with Chlorine, it forms Strontium Chloride (SrCl₂). The balanced equation will be: \[ \mathrm{Sr} + \mathrm{Cl}_{2} \longrightarrow \mathrm{SrCl}_{2} \]
04

Reaction with Phosphorus (P₄)

When Strontium reacts with Phosphorus, it forms Strontium Phosphide (Sr₃P₂). The balanced equation will be: \[ 6 \mathrm{Sr} + 2 \mathrm{P}_{4} \longrightarrow 3 \mathrm{Sr}_{3}\mathrm{P}_{2} \]
05

Reaction with Hydrogen (H₂)

When Strontium reacts with Hydrogen, it forms Strontium Hydride (SrH₂). The balanced equation will be: \[ \mathrm{Sr} + \mathrm{H}_{2} \longrightarrow \mathrm{SrH}_{2} \]
06

Reaction with Water (H₂O)

When Strontium reacts with Water, it forms Strontium Hydroxide (Sr(OH)₂) and Hydrogen gas. The balanced equation will be: \[ \mathrm{Sr} + 2 \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{Sr}(\mathrm{OH})_{2} + \mathrm{H}_{2} \]
07

Reaction with Hydrochloric acid (HCl)

When Strontium reacts with Hydrochloric acid, it forms Strontium Chloride (SrCl₂) and Hydrogen gas. The balanced equation will be: \[ \mathrm{Sr} + 2 \mathrm{HCl} \longrightarrow \mathrm{SrCl}_{2} + \mathrm{H}_{2} \]

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While selenic acid has the formula \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) and thus is directly related to sulfuric acid, telluric acid is best visualized as \(\mathrm{H}_{6} \mathrm{TeO}_{6}\) or \(\mathrm{Te}(\mathrm{OH})_{6}\) a. What is the oxidation state of tellurium in \(\operatorname{Te}(\mathrm{OH})_{6} ?\) b. Despite its structural differences with sulfuric and selenic acid, telluric acid is a diprotic acid with \(\mathrm{p} K_{\mathrm{a}_{1}}=7.68\) and \(\mathrm{p} K_{\mathrm{a}_{2}}=11.29 .\) Telluric acid can be prepared by hydrolysis of tellurium hexafluoride according to the equation $$\operatorname{TeF}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\operatorname{Te}(\mathrm{OH})_{6}(a q)+6 \mathrm{HF}(a q)$$ Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas:$$\operatorname{Te}(s)+3 \mathrm{F}_{2}(g) \longrightarrow \operatorname{Te} \mathrm{F}_{6}(g)$$.If a cubic block of tellurium (density \(=6.240 \mathrm{g} / \mathrm{cm}^{3}\) ) measuring \(0.545 \mathrm{cm}\) on edge is allowed to react with 2.34 L fluorine gas at 1.06 atm and \(25^{\circ} \mathrm{C}\), what is the \(\mathrm{pH}\) of a solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated \(\operatorname{Te} \mathrm{F}_{6}(g)\) in \(115 \mathrm{mL}\) solution? Assume \(100 \%\) yield in all reactions.

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