Chapter 4.3, Problem 4.11BFP

(a)

Interpretation Introduction

Interpretation:

The mass of solid $\text{mercury}\left(\text{II}\right)\text{sulfide}$ that can form when $0.050\text{L}$ of $0.010M\text{mercury}\left(\text{II}\right)\text{nitrate}$ reacts with $0.020\text{L}$ of $0.10M$ sodium sulfide is to be calculated.

Concept introduction:

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction. The amount of product depends on the amount of limiting reagent since the product formation is not possible in the absence of it.

(b)

Interpretation Introduction

Interpretation:

A reaction table for the given reaction is to be written.

Concept introduction:

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction. The amount of product depends on the amount of limiting reagent since the product formation is not possible in the absence of it.

The molecular equation for the reaction between $\text{mercury}\left(\text{II}\right)\text{nitrate}$ and sodium sulfide is:

$\text{Hg}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+{\text{Na}}_2\text{S}\left(aq\right)\to \text{HgS}\left(s\right)+2{\text{NaNO}}_3\left(aq\right)$