Final answer:
According to solubility rules, NaNO3 is soluble due to its nitrate ion and would not precipitate. While sulfides are often insoluble, ammonium sulfide ((NH4)2S) is also typically soluble, and therefore, no solid precipitate is expected for these given reaction products.
Step-by-step explanation:
During a double displacement reaction in an aqueous solution, we assess the solubility rules to determine if a solid precipitate will form. Given the possible products (NH4)2S and NaNO3, we can apply solubility rules to predict which compounds are insoluble.
According to solubility guidelines, nitrates are generally soluble in water. This rule means NaNO3, sodium nitrate, would remain in solution as aqueous ions, and no precipitate would form. On the other hand, sulfides are often insoluble, especially those combined with divalent cations such as Ca2+, Sr2+, and Ba2+, though ammonium sulfide (NH4)2S is generally regarded as soluble as well. In this scenario, we must consider that rules can have exceptions or be conditional based on specific ion combinations. Complete and comprehensive solubility tables should be referenced to confirm the solubility of (NH4)2S in water.
Given the guidelines and typical solubility patterns, we are inclined to assume that, since NaNO3 will remain in solution due to its nitrate ion, and assuming (NH4)2S behaves according to general solubility rules for sulfides, neither of these compounds is likely to form a solid precipitate under ordinary conditions. Therefore, no precipitate is expected to form from these specific reaction products.