Final answer:
The question asks for the mass of precipitate formed from mixing ammonium carbonate and silver nitrate solutions. The reaction produces silver carbonate as the precipitate, but without the full reaction or additional data, the exact mass of the precipitate cannot be calculated.
Step-by-step explanation:
To determine the mass of the precipitate formed when a 50.00 ml sample of 0.250 M ammonium carbonate is mixed with 60.0 ml of a 0.300 M silver nitrate solution, we must first identify the reaction that occurs and which product is the precipitate.
The reaction between ammonium carbonate (NH4)2CO3) and silver nitrate (AgNO3) forms silver carbonate (Ag2CO3) as the precipitate with ammonium nitrate (NH4NO3) remaining in solution:
2 AgNO3 (aq) + (NH4)2CO3 (aq) → Ag2CO3 (s) + 2 NH4NO3 (aq)
To calculate the mass, we need to find out which reactant is the limiting reagent and then use stoichiometry to determine the moles of Ag2CO3 produced. Since stoichiometry is based on mole-to-mole conversions, we'd convert the volumes and concentrations (in molarity) to moles, use the molar ratio from the balanced equation, and finally multiply by the molar mass of Ag2CO3 to get the mass in grams.
However, this question does not provide sufficient data or a full reaction equation to complete the calculation. Therefore, this answer will not include the final mass of the precipitate, as the required information is not fully available.