Chapter 4, Problem 92GQ

Aqueous solutions of iron(II) chloride and sodium sulfide react to form iron(II)sulfide and sodium chloride.

(a) Write the balanced equation for the reaction.

(b) If you combine 40. g each of Na₂S and FeCl₂, what is the limiting reactant?

(c) What mass of FeS is produced?

(d) What mass of Na₂S or FeCl₂ remains after the reaction?

(e) What mass of FeCl₂ is required to react completely with 40. g of Na₂S?

Interpretation Introduction

Interpretation:

Balanced equation for the given reaction has to be written.

Concept introduction:

• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.

Answer to Problem 92GQ

The balanced equation for the reaction of aqueous solutions of iron(II) chloride and sodium sulfate forming iron(II) sulfide is written as,

  $FeC{l}_{2(aq)}+N{a}_2{S}_{(s)}\stackrel{}{\to }Fe{S}_{(s)}+2NaC{l}_{(aq)}$

Explanation of Solution

Aqueous solution of iron (II) chloride and sodium sulfide reacts to form iron (II) sulfide and sodium chloride.

For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.

Balanced equation for this reaction is,

  $FeC{l}_{2(aq)}+N{a}_2{S}_{(s)}\stackrel{}{\to }Fe{S}_{(s)}+2NaC{l}_{(aq)}$

Interpretation Introduction

Interpretation:

Limiting reagent of the reaction of $\text{40}\text{.0 g}$ of ${\text{Na}}_{\text{2}}\text{S}$ with $\text{40}\text{.0 g}$ of ${\text{FeCl}}_{\text{2}}$ has to be determined.

Concept introduction:

Limiting reagent: limiting reagent is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction.

Answer to Problem 92GQ

${\text{FeCl}}_2$ is the limiting reactant

Explanation of Solution

Balanced chemical equation for the given reaction is,

  $FeC{l}_{2(aq)}+N{a}_2{S}_{(s)}\stackrel{}{\to }Fe{S}_{(s)}+2NaC{l}_{(aq)}$

To find the mass of product formed in the reaction, the amount of each reactant involved in the given reaction should be determined.

Therefore,

  $\begin{array}{l}\text{Amount}\text{of}{\text{FeCl}}_{\text{2}}\text{=}\text{40}\text{.0g}{\text{FeCl}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{FeCl}}_{\text{2}}}{\text{126}\text{.75}\text{g}}\\ \text{=}\text{0}\text{.315}\text{mol}{\text{FeCl}}_{\text{2}}\\ \text{Amount}\text{of}{\text{Na}}_{\text{2}}\text{S}\text{=}\text{40}\text{.0g}{\text{Na}}_{\text{2}}\text{S}\text{×}\frac{\text{1}\text{mol}{\text{Na}}_{\text{2}}\text{S}}{\text{78}\text{.05}\text{g}}\\ \text{=}\text{0}\text{.5124}\text{mol}{\text{Na}}_{\text{2}}\text{S}\end{array}$

The mass of $\text{FeS}$ and $\text{NaCl}$ expected based on each reactant is,

  $\begin{array}{l}\text{0}\text{.315}\text{mol}{\text{FeCl}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}\text{FeS}\text{formed}}{\text{1}\text{mol}{\text{FeCl}}_{\text{2}}\text{available}}\text{×}\frac{\text{87}\text{.91}\text{g}\text{FeS}}{\text{1}\text{mol}\text{FeS}}\text{=}\text{27}\text{.69}\text{g}\text{FeS}\\ \\ \text{0}\text{.5124}\text{mol}{\text{Na}}_{\text{2}}\text{S}\text{×}\frac{\text{2}\text{mol}\text{NaClformed}}{\text{1}\text{mol}{\text{Na}}_{\text{2}}\text{S}\text{available}}\text{×}\frac{\text{58}\text{.44}\text{g}\text{NaCl}}{\text{1}\text{mol}\text{NaCl}}\text{=}\text{59}\text{.88g}\text{NaCl}\end{array}$

The amount of ${\text{FeCl}}_2$ available produces products in lower quantity than the amount of ${\text{Na}}_{\text{2}}\text{S}$ available.

Therefore ${\text{FeCl}}_2$ is the limiting reactant

Interpretation Introduction

Interpretation:

Mass of Iron produced in the reaction of $\text{40}\text{.0 g}$ of ${\text{Na}}_{\text{2}}\text{S}$ with $\text{40}\text{.0 g}$ of ${\text{FeCl}}_{\text{2}}$ has to be calculated.

Answer to Problem 92GQ

The mass of $\text{FeS}$ produced is $\text{27}\text{.69}\text{g}$.

Explanation of Solution

As calculated in previous step, the maximum mass of $\text{FeS}$ that can be produced is $\text{27}\text{.69}\text{g}$.

Interpretation Introduction

Interpretation:

The mass of ${\text{Na}}_2\text{S}$ remains after reaction has to be calculated.

Answer to Problem 92GQ

The mass of ${\text{Na}}_2\text{S}$ remains after reaction is $\text{15}\text{.42}\text{g}$

Explanation of Solution

The mass of ${\text{Na}}_{\text{2}}\text{S}$ remains when all the ${\text{FeCl}}_2$ has been converted to product can be determined as follows,

The mass of ${\text{Na}}_{\text{2}}\text{S}$ required to react with all the ${\text{FeCl}}_2$ is,

  $\text{0}\text{.315}\text{mol}{\text{FeCl}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{Na}}_{\text{2}}\text{S}}{\text{1mol}{\text{FeCl}}_{\text{2}}}\text{×}\frac{\text{78}\text{.05}\text{g}{\text{Na}}_{\text{2}}\text{S}}{\text{1}\text{mol}{\text{Na}}_{\text{2}}\text{S}}\text{=}\text{24}\text{.58}\text{g}{\text{Na}}_{\text{2}}\text{S}\text{required}$

Reaction began with $\text{40}\text{.0}\text{g}$ of ${\text{Na}}_{\text{2}}\text{S}$ but only $\text{24}\text{.58}\text{g}{\text{Na}}_{\text{2}}\text{S}$ is required as calculated in previous step.  Thus the mass of ${\text{Na}}_{\text{2}}\text{S}$ that remains after the reaction is,

  $\text{40}\text{.0}\text{g}{\text{Na}}_{\text{2}}\text{S}\text{present}\text{-}\text{24}\text{.58}\text{g}{\text{Na}}_{\text{2}}\text{S}\text{required}\text{=}\text{15}\text{.42}\text{g}{\text{Na}}_{\text{2}}\text{S}\text{left}$

Interpretation Introduction

Interpretation:

The mass of ${\text{FeCl}}_2$ required to react completely with $\text{40}\text{.0}\text{g}\text{of}{\text{Na}}_{\text{2}}\text{S}$

Concept Introduction:

Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.

Explanation of Solution

Balanced chemical equation for the given reaction is,

  $FeC{l}_{2(aq)}+N{a}_2{S}_{(s)}\stackrel{}{\to }Fe{S}_{(s)}+2NaC{l}_{(aq)}$

From the balanced equation it is clear that the reactants $FeC{l}_{2}andN{a}_{2}S$ reacts in $1:1ratio.$

The mass of ${\text{FeCl}}_2$ required to react completely with $\text{40}\text{.0}\text{g}\text{of}{\text{Na}}_{\text{2}}\text{S}$ can be calculated as follows,

  $\text{Amount}\text{of}{\text{Na}}_{\text{2}}\text{S}\text{=}\text{40}\text{.0g}{\text{Na}}_{\text{2}}\text{S}\text{×}\frac{\text{1}\text{mol}{\text{Na}}_{\text{2}}\text{S}}{\text{78}\text{.05}\text{g}}\text{=}\text{0}\text{.5124}\text{mol}{\text{Na}}_{\text{2}}\text{S}$

Mass of ${\text{FeCl}}_2$ required to react completely with $\text{40}\text{.0}\text{g}\text{of}{\text{Na}}_{\text{2}}\text{S}$ is,

  $\text{0}\text{.5124}\text{mol}{\text{Na}}_{\text{2}}\text{S}\text{×}\frac{\text{1}\text{mol}{\text{FeCl}}_{\text{2}}}{\text{1}\text{mol}{\text{Na}}_{\text{2}}\text{S}}\text{×}\frac{\text{126}\text{.75}\text{g}{\text{FeCl}}_{\text{2}}}{\text{1}\text{mol}{\text{FeCl}}_{\text{2}}}\text{=}\text{64}\text{.9467}\text{g}{\text{FeCl}}_{\text{2}}$