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Chapter 19: Problem 38

Write equations describing the reactions of Sn with each of the following: \(\mathrm{Cl}_{2}, \mathrm{O}_{2},\) and HCl.

Short Answer

Expert verified
Reaction of Tin with Chlorine Gas: \(Sn + 2 Cl_2 \rightarrow SnCl_4\) Reaction of Tin with Oxygen Gas: \(2 Sn + O_2 \rightarrow 2 SnO\) (tin (II) oxide) \(Sn + O_2 \rightarrow SnO_2\) (tin (IV) oxide) Reaction of Tin with Hydrochloric Acid: \(Sn + 2 HCl \rightarrow SnCl_2 + H_2\)

Step by step solution

01

Reaction of Tin with Chlorine Gas

To find the chemical formula for tin (IV) chloride, we combine Sn with 4 chlorine (Cl) atoms. This creates the compound SnCl₄. Now we balance the equation: Sn (s) + 2 Cl₂ (g) -> SnCl₄ (s) Balanced Equation: Sn + 2 Cl₂ -> SnCl₄
02

Reaction of Tin with Oxygen Gas

Tin forms two oxides: tin (II) oxide (SnO) and tin (IV) oxide (SnO₂). In both cases, Sn is combined with oxygen (O) atoms. In tin (II) oxide, tin has a +2 oxidation state, while in tin (IV) oxide, tin has a +4 oxidation state. We need to balance the equations for both reactions: For tin (II) oxide: 2 Sn (s) + O₂ (g) -> 2 SnO (s) Balanced Equation: 2 Sn + O₂ -> 2 SnO For tin (IV) oxide: Sn (s) + O₂ (g) -> SnO₂ (s) Balanced Equation: Sn + O₂ -> SnO₂
03

Reaction of Tin with Hydrochloric Acid

When tin reacts with hydrochloric acid (HCl), tin (II) chloride (SnCl₂) is formed along with hydrogen gas (H₂). We combine Sn with 2 chlorine (Cl) atoms and 2 hydrogen (H) atoms to create the products SnCl₂ and H₂. Now we balance the equation: Sn (s) + 2 HCl (aq) -> SnCl₂ (aq) + H₂ (g) Balanced Equation: Sn + 2 HCl -> SnCl₂ + H₂

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Most popular questions from this chapter

Use bond energies to estimate the maximum wavelength of light that will cause the reaction $$\mathrm{O}_{3} \stackrel{h v}{\longrightarrow} \mathrm{O}_{2}+\mathrm{O}$$

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