Question

In the production of silver chloride, 25.00 mL of 0.15 M NiCl₂ is reacted with 30.0 mL of 0.15 M AgNO₃.

NiCl₂(aq) + 2 AgNO₃(aq) 2 AgCl(s) + Ni(NO₃)₂(aq)
A. What is the limiting reactant?
B. What mass of precipitate will result?

Answer 1

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.