Final answer:
NiCl₂ is the limiting reactant, and the mass of the AgCl precipitate formed from the reaction is 0.537 grams.
Step-by-step explanation:
Limiting Reactant and Mass of Precipitate
In the reaction between NiCl₂ and AgNO₋, to produce silver chloride (AgCl), we must first determine the limiting reactant. We do this by calculating the number of moles of each reactant:
For NiCl₂: 25.00 mL * 0.15 M = 0.00375 moles
For AgNO₋: 30.0 mL * 0.15 M = 0.00450 moles
According to the balanced equation, NiCl₂ reacts with 2 moles of AgNO₋ for every mole of NiCl₂. Thus, we only require 0.00750 moles of AgNO₋ to react with the available NiCl₂. This means NiCl₂ is the limiting reactant.
To find the mass of silver chloride precipitate formed, we first calculate the moles of AgCl produced:
Moles of AgCl = Moles of NiCl₂ (since the reaction is 1:1 for NiCl₂ and AgCl)
Moles of AgCl = 0.00375 moles
Next, we convert moles of AgCl to mass:
Molar mass of AgCl = 143.32 g/mol
Mass of AgCl = 0.00375 moles * 143.32 g/mol = 0.537 grams
So, the mass of the precipitate formed is 0.537 grams of AgCl.