A 3.50 g of an alloy which contains only lead and tin is dissolved in hot HNO3. Excess sulfuric acid is added to this solution and 1.57 g of PbSO4(s) is obtained. (a) Write the net ionic equation for the formation of PbSO4.

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Identify the ions involved in the formation of PbSO4. Lead(II) ions (Pb^{2+}) and sulfate ions (SO_4^{2-}) are the key players in this reaction.

Write the dissociation equations for the reactants: Pb^{2+} (aq) and SO_4^{2-} (aq) are already in their ionic forms.

Combine the ions to form the solid product: Pb^{2+} (aq) + SO_4^{2-} (aq) \rightarrow PbSO_4 (s).

Ensure the equation is balanced. In this case, the charges and atoms are balanced as written.

The net ionic equation for the formation of lead(II) sulfate is: Pb^{2+} (aq) + SO_4^{2-} (aq) \rightarrow PbSO_4 (s).

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Net Ionic Equations

A net ionic equation represents the chemical species that are involved in a reaction, excluding spectator ions that do not participate. It focuses on the actual chemical change occurring in the solution, providing a clearer picture of the reaction. In this case, the formation of PbSO4 from lead ions and sulfate ions will be highlighted.

Solubility Rules

Solubility rules help predict whether a compound will dissolve in water or form a precipitate. Lead(II) sulfate (PbSO4) is known to be insoluble in water, which means it will precipitate out of the solution when lead ions react with sulfate ions. Understanding these rules is essential for determining the products of the reaction.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced equations. In this scenario, knowing the mass of PbSO4 produced allows for the determination of the amount of lead that reacted, which is crucial for understanding the composition of the original alloy and the efficiency of the reaction.

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