Question

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

Short answer

The balanced chemical equations for the reactions between tin (Sn) and the given substances are as follows: 1. Reaction with Chlorine gas (Cl₂): \( Sn + 2Cl_2 \rightarrow SnCl_4 \) 2. Reaction with Oxygen gas (O₂): \( Sn + O_2 \rightarrow SnO_2 \) 3. Reaction with Hydrochloric acid (HCl): \( Sn + 2HCl \rightarrow SnCl_2 + H_2 \)

Step-by-step solution

Reaction with Chlorine gas (Cl₂)

: When tin (Sn) reacts with chlorine gas (Cl₂), it forms tin(IV) chloride (SnCl₄). Unbalanced equation: Sn + Cl₂ → SnCl₄ To balance the equation, we need 2 Cl₂ molecules for every Sn atom: Balanced equation: Sn + 2Cl₂ → SnCl₄

Reaction with Oxygen gas (O₂)

: When tin reacts with oxygen gas, it forms tin(IV) oxide (SnO₂). Unbalanced equation: Sn + O₂ → SnO₂ The equation is already balanced: Balanced equation: Sn + O₂ → SnO₂

Reaction with Hydrochloric acid (HCl)

: Tin reacts with hydrochloric acid to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂). Unbalanced equation: Sn + HCl → SnCl₂ + H₂ To balance the equation, we need 2 HCl molecules for every Sn atom: Balanced equation: Sn + 2HCl → SnCl₂ + H₂

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial skill in chemistry that ensures the law of conservation of mass is upheld. In a balanced chemical equation, the number of each type of atom on the reactants side must be equal to the number of each type of atom on the products side. This is because atoms are not created or destroyed in a chemical reaction—they are simply rearranged.

For instance, when balancing the reaction of tin (Sn) with chlorine gas (Cl₂), we start with the unbalanced equation Sn + Cl₂ → SnCl₄. Upon inspection, we notice that we have one tin atom on both sides, but the chlorine atoms are not balanced. There are two chlorine atoms in the reactant Cl₂ and four in the product SnCl₄. To balance this, we must double the amount of chlorine gas in the reactants, which gives us Sn + 2Cl₂ → SnCl₄, now a balanced equation. Similarly, when balancing the reaction with hydrochloric acid, we must ensure that the number of hydrogen and chlorine atoms are equal on both sides of the equation.

Reaction with Chlorine

When tin (Sn) reacts with chlorine gas (Cl₂), a synthesis reaction occurs, and tin(IV) chloride (SnCl₄) is formed. This is an example of a halogenation process since chlorine is a halogen. The balanced chemical equation for this reaction is Sn + 2Cl₂ → SnCl₄.

This chemical reaction is important in the production of tin compounds used in organotin chemistry, which can be applied to various industrial processes. Understanding the specific stoichiometry, or the ratio of how much chlorine reacts with tin, helps chemists prepare the correct amounts of material for large scale reactions without waste.

Reaction with Oxygen

Tin's reaction with oxygen (O₂) is a simple synthesis reaction where oxygen combines with tin to form tin(IV) oxide (SnO₂). The equation representing this reaction is Sn + O₂ → SnO₂, which is already balanced as written, with one tin atom and two oxygen atoms on both sides.

Tin(IV) oxide is a crucial compound with several applications, including use as a protective coating for other metals and as an additive in making glass. The reaction is straightforward, but still plays a significant role in the materials industry. It's a great example of how even the simplest balanced reactions can be incredibly valuable.

Reaction with Hydrochloric Acid

When tin reacts with hydrochloric acid (HCl), it produces tin(II) chloride (SnCl₂) and hydrogen gas (H₂). The unbalanced chemical equation for the process is Sn + HCl → SnCl₂ + H₂. To balance this, we recognize that for every atom of tin, two molecules of HCl are required. This leads to a balanced equation of Sn + 2HCl → SnCl₂ + H₂.

This reaction is a type of single displacement reaction, where an element displaces another in a compound. In this context, the reaction is also important for refining and utilizing tin in various industrial processes, including metal plating and as a catalyst in certain organic reactions. It's essential to understand both the stoichiometry and the type of reaction taking place to accurately predict the products and their quantities.