Chapter 6, Problem 41QAP

Balance each of the following chemical equations.

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msp;  ${\text{Fe}}_3{\text{O}}_4\left(s\right)+{\text{H}}_2\left(g\right)\to \text{Fe}\left(l\right)+{\text{H}}_2\text{O}\left(g\right)$

i>   ${\text{K}}_2{\text{SO}}_4\left(aq\right)+{\text{BaCl}}_2\left(aq\right)\to {\text{BaSO}}_4\left(s\right)+\text{KCl}\left(aq\right)$

i>   $\text{HCl}\left(aq\right)+\text{FeS}\left(s\right)\to {\text{FeCl}}_2\left(aq\right)+{\text{H}}_2\text{S}\left(g\right)$

i>   ${\text{Br}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{SO}}_2\left(g\right)\to \text{HBr}\left(aq\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)$

i>   ${\text{CS}}_2\left(l\right)+{\text{Cl}}_2\left(g\right)\to {\text{CCl}}_4\left(l\right)+{\text{S}}_2{\text{Cl}}_2\left(g\right)$

i>   ${\text{Cl}}_2{\text{O}}_7\left(g\right)+\text{Ca}{\left(\text{OH}\right)}_2\left(aq\right)\to \text{Ca}{\left({\text{ClO}}_4\right)}_2\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

i>   ${\text{PBr}}_3\left(l\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_3{\text{PO}}_3\left(aq\right)+\text{HBr}\left(g\right)$

i>   $\text{Ba}{\left({\text{ClO}}_3\right)}_2\left(s\right)\to {\text{BaCl}}_2\left(s\right)+{\text{O}}_2\left(s\right)$

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Interpretation Introduction

(a)

Interpretation:

The given chemical equation is to be balanced.

Concept Introduction:

In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.

Answer to Problem 41QAP

$\text{F}{\text{e}}_3{\text{O}}_4(s)+4{\text{H}}_2(g)\to 3\text{F}\text{e}(l)+4{\text{H}}_2\text{O}(g)$.

Explanation of Solution

The provided reaction is:

$\text{F}{\text{e}}_3{\text{O}}_4(s)+{\text{H}}_2(g)\to \text{F}\text{e}(l)+{\text{H}}_2\text{O}(g)$

The most complicated molecule is not clear, so we start arbitrarily. If we place a coefficient of 3 before Fe and 4 before H₂ and H₂ O, we get a balanced equation.

$\text{F}{\text{e}}_3{\text{O}}_4(s)+4{\text{H}}_2(g)\to 3\text{F}\text{e}(l)+4{\text{H}}_2\text{O}(g)$.

Interpretation Introduction

(b)

Interpretation:

The given chemical equation is to be balanced.

Concept Introduction:

In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.

Answer to Problem 41QAP

${\text{K}}_2\text{S}{\text{O}}_4(aq)+\text{B}\text{a}\text{C}{\text{l}}_2(aq)\to \text{B}\text{a}\text{S}{\text{O}}_4(s)+\text{K}\text{C}\text{l}(aq)$.

Explanation of Solution

The given unbalanced equation is:

${\text{K}}_2\text{S}{\text{O}}_4(aq)+\text{B}\text{a}\text{C}{\text{l}}_2(aq)\to \text{B}\text{a}\text{S}{\text{O}}_4(s)+\text{K}\text{C}\text{l}(aq)$

We start with placing 2 before KCl and we get the balanced chemical reaction as follows:

${\text{K}}_2\text{S}{\text{O}}_4(aq)+\text{B}\text{a}\text{C}{\text{l}}_2(aq)\to \text{B}\text{a}\text{S}{\text{O}}_4(s)+2\text{K}\text{C}\text{l}(aq)$.

Interpretation Introduction

(c)

Interpretation:

The given chemical equation is to be balanced.

Concept Introduction:

In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.

Answer to Problem 41QAP

$2\text{H}\text{C}\text{l}(aq)+\text{F}\text{e}\text{S}(s)\to \text{F}\text{e}\text{C}{\text{l}}_2(aq)+{\text{H}}_2\text{S}(g)$.

Explanation of Solution

The given unbalanced equation is:

$\text{H}\text{C}\text{l}(aq)+\text{F}\text{e}\text{S}(s)\to \text{F}\text{e}\text{C}{\text{l}}_2(aq)+{\text{H}}_2\text{S}(g)$

We start by balancing HCl. If we place a coefficient 2 before HCl we get a balanced chemical reaction as follows:

$2\text{H}\text{C}\text{l}(aq)+\text{F}\text{e}\text{S}(s)\to \text{F}\text{e}\text{C}{\text{l}}_2(aq)+{\text{H}}_2\text{S}(g)$.

Interpretation Introduction

(d)

Interpretation:

The given chemical equation is to be balanced.

Concept Introduction:

In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.

Answer to Problem 41QAP

$\text{B}{\text{r}}_2(g)+2{\text{H}}_2\text{O}(l)+\text{S}{\text{O}}_2(g)\to 2\text{H}\text{B}\text{r}(aq)+{\text{H}}_2\text{S}{\text{O}}_4(aq)$.

Explanation of Solution

The provided equation is:

$\text{B}{\text{r}}_2(g)+{\text{H}}_2\text{O}(l)+\text{S}{\text{O}}_2(g)\to \text{H}\text{B}\text{r}(aq)+{\text{H}}_2\text{S}{\text{O}}_4(aq)$

If we place a coefficient of 2 H₂ O and HBr, we get a balanced equation as follows:

$\text{B}{\text{r}}_2(g)+2{\text{H}}_2\text{O}(l)+\text{S}{\text{O}}_2(g)\to 2\text{H}\text{B}\text{r}(aq)+{\text{H}}_2\text{S}{\text{O}}_4(aq)$.

Interpretation Introduction

(e)

Interpretation:

The given chemical equation is to be balanced.

Concept Introduction:

In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.

Answer to Problem 41QAP

$\text{C}{\text{S}}_2(l)+3\text{C}{\text{l}}_2(g)\to \text{C}\text{C}{\text{l}}_4(l)+{\text{S}}_2\text{C}{\text{l}}_2(g)$.

Explanation of Solution

The provided equation is:

$\text{C}{\text{S}}_2(l)+\text{C}{\text{l}}_2(g)\to \text{C}\text{C}{\text{l}}_4(l)+{\text{S}}_2\text{C}{\text{l}}_2(g)$

If we place a coefficient of 3 before Cl₂, we get the balanced chemical reaction as follows:

$\text{C}{\text{S}}_2(l)+3\text{C}{\text{l}}_2(g)\to \text{C}\text{C}{\text{l}}_4(l)+{\text{S}}_2\text{C}{\text{l}}_2(g)$.

Interpretation Introduction

(f)

Interpretation:

The given chemical equation is to be balanced.

Concept Introduction:

In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.

Answer to Problem 41QAP

$\text{C}{\text{l}}_2{\text{O}}_7(g)+\text{C}\text{a}{(\text{O}\text{H})}_2(aq)\to \text{C}\text{a}{(\text{C}\text{l}{\text{O}}_4)}_2(aq)+{\text{H}}_2\text{O}(l)$.

Explanation of Solution

The provided reaction is:

$\text{C}{\text{l}}_2{\text{O}}_7(g)+\text{C}\text{a}{(\text{O}\text{H})}_2(aq)\to \text{C}\text{a}{(\text{C}\text{l}{\text{O}}_4)}_2(aq)+{\text{H}}_2\text{O}(l)$

We can see that all the atoms on both the sides are already balanced. Therefore, we already have a balanced equation.

$\text{L}{\text{i}}_2\text{O}(s)+{\text{H}}_2\text{O}(l)\to 2\text{L}\text{i}\text{O}\text{H}(aq)$.

Interpretation Introduction

(g)

Interpretation:

The given chemical equation is to be balanced.

Concept Introduction:

In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.

Answer to Problem 41QAP

$\text{P}\text{B}{\text{r}}_3(l)+3{\text{H}}_2\text{O}(l)\to {\text{H}}_3\text{P}{\text{O}}_4(aq)+3\text{H}\text{B}\text{r}(g)$.

Explanation of Solution

The provided reaction is:

$\text{P}\text{B}{\text{r}}_3(l)+{\text{H}}_2\text{O}(l)\to {\text{H}}_3\text{P}{\text{O}}_4(aq)+\text{H}\text{B}\text{r}(g)$

The most complicated molecule is ${\text{H}}_3\text{P}{\text{O}}_4$ because it contains the most number of elements (three). We start with the elements of ${\text{H}}_3\text{P}{\text{O}}_4$. We first start with oxygen. To balance oxygen, we place a coefficient 3 before H₂ O

$\text{P}\text{B}{\text{r}}_3(l)+3{\text{H}}_2\text{O}(l)\to {\text{H}}_3\text{P}{\text{O}}_4(aq)+\text{H}\text{B}\text{r}(g)$

We can balance hydrogen by giving HBr a coefficient of 3. We now have a balanced equation

$\text{P}\text{B}{\text{r}}_3(l)+3{\text{H}}_2\text{O}(l)\to {\text{H}}_3\text{P}{\text{O}}_4(aq)+3\text{H}\text{B}\text{r}(g)$.

Interpretation Introduction

(h)

Interpretation:

The given chemical equation is to be balanced.

Concept Introduction:

In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.

Answer to Problem 41QAP

$\text{B}\text{a}{(\text{C}\text{l}{\text{O}}_3)}_2(s)\to \text{B}\text{a}\text{C}{\text{l}}_2(s)+3{\text{O}}_2(g)$.

Explanation of Solution

The provided reaction is:

$\text{B}\text{a}{(\text{C}\text{l}{\text{O}}_3)}_2(s)\to \text{B}\text{a}\text{C}{\text{l}}_2(s)+{\text{O}}_2(g)$

$\text{B}\text{a}{(\text{C}\text{l}{\text{O}}_3)}_2$ is the most complicated element. It contains the most number of elements (three). We arbitrarily start with oxygen. If we place a coefficient of 3 before O₂, we have a balanced equation.

$\text{B}\text{a}{(\text{C}\text{l}{\text{O}}_3)}_2(s)\to \text{B}\text{a}\text{C}{\text{l}}_2(s)+3{\text{O}}_2(g)$.