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 of tin (Sn) with chlorine gas (Cl₂), oxygen gas (O₂), and hydrochloric acid (HCl) are as follows: \( \mathrm{Sn} + 2 \, \mathrm{Cl}_{2} \rightarrow \mathrm{SnCl}_{4} \) \( \mathrm{4 \, Sn} + 2 \, \mathrm{O}_{2} \rightarrow 4 \, \mathrm{SnO}_{2} \) \( \mathrm{Sn} + 2 \, \mathrm{HCl} \rightarrow \mathrm{SnCl}_{2} + \mathrm{H}_{2} \)

Step-by-step solution

Reaction of tin with chlorine gas

The reaction of tin (Sn) with chlorine gas (Cl₂) forms tin(IV) chloride (SnCl₄). To balance the equation, simply combine the elements as shown below: \[\mathrm{Sn} + 2 \, \mathrm{Cl}_{2} \rightarrow \mathrm{SnCl}_{4}\]

Reaction of tin with oxygen gas

Tin reacts with oxygen gas (O₂) to form tin(IV) oxide (SnO₂). To balance the equation, we need to adjust the coefficients for the reactants and products, which results in the following balanced equation: \[\mathrm{4 \, Sn} + 2 \, \mathrm{O}_{2} \rightarrow 4 \, \mathrm{SnO}_{2}\]

Reaction of tin with hydrochloric acid

Tin reacts with hydrochloric acid (HCl) to form tin(II) chloride (SnCl₂) and hydrogen gas (H₂). To balance the equation, we will adjust the coefficients to ensure the same number of atoms for each element are on both sides: \[\mathrm{Sn} + 2 \, \mathrm{HCl} \rightarrow \mathrm{SnCl}_{2} + \mathrm{H}_{2}\] Now we have successfully written and balanced the chemical equations for the reactions of tin with chlorine gas, oxygen gas, and hydrochloric acid.