Chapter 6, Problem 42QAP

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 42QAP

$4\text{N}\text{a}\text{C}\text{l}(s)+2\text{S}{\text{O}}_2(g)+2{\text{H}}_2\text{O}(g)+{\text{O}}_2(g)\to 2\text{N}{\text{a}}_2\text{S}{\text{O}}_4(s)+4\text{H}\text{C}\text{l}(g)$.

Explanation of Solution

The provided reaction is:

$\text{N}\text{a}\text{C}\text{l}(s)+\text{S}{\text{O}}_2(g)+{\text{H}}_2\text{O}(g)+{\text{O}}_2(g)\to \text{N}{\text{a}}_2\text{S}{\text{O}}_4(s)+\text{H}\text{C}\text{l}(g)$

The most complicated molecule is $\text{N}{\text{a}}_2\text{S}{\text{O}}_4$ because it contains the most number of elements. If we place a coefficient of 2 before NaCl and a coefficient of 2 before H₂ O, SO₂ and $\text{N}{\text{a}}_2\text{S}{\text{O}}_4$

$2\text{N}\text{a}\text{C}\text{l}(s)+2\text{S}{\text{O}}_2(g)+2{\text{H}}_2\text{O}(g)+{\text{O}}_2(g)\to 2\text{N}{\text{a}}_2\text{S}{\text{O}}_4(s)+\text{H}\text{C}\text{l}(g)$

Finally, we can balance Na, H and Cl by giving both NaCl and HCl a coefficient of 4. We get a balanced equation.

$4\text{N}\text{a}\text{C}\text{l}(s)+2\text{S}{\text{O}}_2(g)+2{\text{H}}_2\text{O}(g)+{\text{O}}_2(g)\to 2\text{N}{\text{a}}_2\text{S}{\text{O}}_4(s)+4\text{H}\text{C}\text{l}(g)$.

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 42QAP

$3\text{B}{\text{r}}_2(l)+{\text{I}}_2(s)\to 2\text{I}\text{B}{\text{r}}_3(s)$.

Explanation of Solution

The given unbalanced equation is:

$\text{B}{\text{r}}_2(l)+{\text{I}}_2(s)\to \text{I}\text{B}{\text{r}}_3(s)$

We start with elements of $\text{I}\text{B}{\text{r}}_3$. To balance Br, we need to place 2 before $\text{I}\text{B}{\text{r}}_3$ and 3 before Br₂. Thus, the balanced chemical reaction is as follows:

$3\text{B}{\text{r}}_2(l)+{\text{I}}_2(s)\to 2\text{I}\text{B}{\text{r}}_3(s)$.

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 42QAP

$\text{C}{\text{a}}_3{\text{N}}_2(s)+6{\text{H}}_2\text{O}(l)\to 3\text{C}\text{a}{(\text{O}\text{H})}_2(aq)+2\text{N}{\text{H}}_3(g)$.

Explanation of Solution

The given unbalanced equation is:

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

We start by balancing Ca(OH)₂, if we place a coefficient 3 before Ca(OH)₂, and 6 before H₂ O

$\text{C}{\text{a}}_3{\text{N}}_2(s)+6{\text{H}}_2\text{O}(l)\to 3\text{C}\text{a}{(\text{O}\text{H})}_2(aq)+\text{N}{\text{H}}_3(g)$

Finally, to balance N and H, we place a coefficient of 2 before NH₃. Thus, the balanced chemical reaction is as follows:

$\text{C}{\text{a}}_3{\text{N}}_2(s)+6{\text{H}}_2\text{O}(l)\to 3\text{C}\text{a}{(\text{O}\text{H})}_2(aq)+2\text{N}{\text{H}}_3(g)$.

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 42QAP

$2\text{B}{\text{F}}_3(g)+3{\text{H}}_2\text{O}(g)\to {\text{B}}_2{\text{O}}_3(s)+6\text{H}\text{F}(g)$.

Explanation of Solution

The provided equation is:

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

If we place coefficients 2 before BF₃, 3 before H₂ O and 6 before HF, we get a balanced equation.

$2\text{B}{\text{F}}_3(g)+3{\text{H}}_2\text{O}(g)\to {\text{B}}_2{\text{O}}_3(s)+6\text{H}\text{F}(g)$.

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 42QAP

$\text{S}{\text{O}}_2(g)+2\text{C}{\text{l}}_2(g)\to \text{S}\text{O}\text{C}{\text{l}}_2(l)+\text{C}{\text{l}}_2\text{O}(g)$.

Explanation of Solution

The provided equation is:

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

The most complicated molecule is $\text{S}\text{O}\text{C}{\text{l}}_2$ because it contains the most number of elements. If we place a coefficient of 2 before Cl₂, we get the balanced chemical reaction as follows:

$\text{S}{\text{O}}_2(g)+2\text{C}{\text{l}}_2(g)\to \text{S}\text{O}\text{C}{\text{l}}_2(l)+\text{C}{\text{l}}_2\text{O}(g)$.

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 42QAP

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

Explanation of Solution

The provided reaction is:

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

Start with most complicated element $\text{L}\text{i}\text{O}\text{H}$. We need to place a coefficient of 2 before $\text{L}\text{i}\text{O}\text{H}$. Now, we 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 42QAP

$\text{M}\text{g}(s)+\text{C}\text{u}\text{O}(s)\to \text{M}\text{g}\text{O}(s)+\text{C}\text{u}(l)$.

Explanation of Solution

The provided reaction is:

$\text{M}\text{g}(s)+\text{C}\text{u}\text{O}(s)\to \text{M}\text{g}\text{O}(s)+\text{C}\text{u}(l)$

All the atoms are balanced. Hence, the equation is balanced.

$\text{M}\text{g}(s)+\text{C}\text{u}\text{O}(s)\to \text{M}\text{g}\text{O}(s)+\text{C}\text{u}(l)$.

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 42QAP

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

Explanation of Solution

The provided reaction is:

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

If we place a coefficient of 3 before Fe, we get:

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

Now, if we place a coefficient of 4 before H₂ and H₂ O, we have a balanced equation.

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