Chapter 6, Problem 44QAP

Interpretation Introduction

(a)

Interpretation:

A balanced equation of the given reaction should be determined.

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 present before the compounds are changed but not the subscripts in the formulas.

Answer to Problem 44QAP

$\text{B}\text{a}{(\text{N}{\text{O}}_3)}_2(aq)+\text{N}{\text{a}}_2\text{C}\text{r}{\text{O}}_4(aq)\to \text{B}\text{a}\text{C}\text{r}{\text{O}}_4(s)+2\text{N}\text{a}\text{N}{\text{O}}_3(aq)$.

Explanation of Solution

The following unbalanced equation is given:

$\text{B}\text{a}{(\text{N}{\text{O}}_3)}_2(aq)+\text{N}\text{a}\text{C}\text{r}{\text{O}}_4(aq)\to \text{B}\text{a}\text{C}\text{r}{\text{O}}_4(s)+\text{N}\text{a}\text{N}{\text{O}}_3(aq)$

There is no clear choice for the most complicated molecule. We arbitrarily start with the elements of BaNO₃ To balance oxygen, we place a coefficient of 2 before NaNO₃. We have got a balanced equation.

$\text{B}\text{a}{(\text{N}{\text{O}}_3)}_2(aq)+\text{N}{\text{a}}_2\text{C}\text{r}{\text{O}}_4(aq)\to \text{B}\text{a}\text{C}\text{r}{\text{O}}_4(s)+2\text{N}\text{a}\text{N}{\text{O}}_3(aq)$.

Interpretation Introduction

(b)

Interpretation:

A balanced equation of the given reaction should be determined.

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 present before the compounds are changed but not the subscripts in the formulas.

Answer to Problem 44QAP

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

Explanation of Solution

The following unbalanced equation is given:

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

There is no clear choice for the most complicated molecule. We arbitrarily start with the elements of PbCl₂ To balance chlorine, we place a coefficient of 2 before KCl. We have got a balanced equation.

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

Interpretation Introduction

(c)

Interpretation:

A balanced equation of the given reaction should be determined.

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 present before the compounds are changed but not the subscripts in the formulas.

Answer to Problem 44QAP

${\text{C}}_2{\text{H}}_5\text{O}\text{H}(l)+3{\text{O}}_2(g)\to 2\text{C}{\text{O}}_2(g)+3{\text{H}}_2\text{O}(l)$.

Explanation of Solution

The following unbalanced equation is given:

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

The most complicated molecule is C₂ H₅ OH because it contains the most elements (three). To balance C, we place a coefficient ot 2 before CO₂.

${\text{C}}_2{\text{H}}_5\text{O}\text{H}(l)+{\text{O}}_2(g)\to 2\text{C}{\text{O}}_2(g)+{\text{H}}_2\text{O}(l)$

To balance H, we place a coefficient of 3 before H₂ O.

${\text{C}}_2{\text{H}}_5\text{O}\text{H}(l)+{\text{O}}_2(g)\to 2\text{C}{\text{O}}_2(g)+3{\text{H}}_2\text{O}(l)$

Finally, to balance O, we place a coefficient of 3 before O₂.

${\text{C}}_2{\text{H}}_5\text{O}\text{H}(l)+3{\text{O}}_2(g)\to 2\text{C}{\text{O}}_2(g)+3{\text{H}}_2\text{O}(l)$

We now have a balanced equation.

Interpretation Introduction

(d)

Interpretation:

A balanced equation of the given reaction should be determined.

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 present before the compounds are changed but not the subscripts in the formulas.

Answer to Problem 44QAP

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

Explanation of Solution

The following unbalanced equation is given:

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

The most complicated molecule is Ca(OH)₂ because it contains the most elements (three). We start with hydrogen. To balance H we place a coefficient of 2 before H₂ O. We now have a balanced equation.

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

Interpretation Introduction

(e)

Interpretation:

A balanced equation of the given reaction should be determined.

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 present before the compounds are changed but not the subscripts in the formulas.

Answer to Problem 44QAP

$\text{S}\text{r}(s)+2\text{H}\text{N}{\text{O}}_3(aq)\to \text{S}\text{r}{(\text{N}{\text{O}}_3)}_2(aq)+{\text{H}}_2(g)$.

Explanation of Solution

The following unbalanced equation is given:

$\text{S}\text{r}(s)+\text{H}\text{N}{\text{O}}_3(aq)\to \text{S}\text{r}\text{N}{\text{O}}_3(aq)+{\text{H}}_2(g)$

There is no clear choice for the most complicated molecule. We arbitrarily start with the elements of SrNO₃ To balance oxygen we, place a coefficient ot 2 before HNO₃. We have got a balanced equation.

$\text{S}\text{r}(s)+2\text{H}\text{N}{\text{O}}_3(aq)\to \text{S}\text{r}{(\text{N}{\text{O}}_3)}_2(aq)+{\text{H}}_2(g)$.

(f)

Interpretation Introduction

Interpretation:

A balanced equation of the given reaction should be determined.

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 present before the compounds are changed but not the subscripts in the formulas.

Answer to Problem 44QAP

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

Explanation of Solution

The following equation is given:

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

It can be seen that all the elements are already balanced. Therefore, we have got a balanced equation.

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

Interpretation Introduction

(g)

Interpretation:

A balanced equation of the given reaction should be determined.

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 present before the compounds are changed but not the subscripts in the formulas.

Answer to Problem 44QAP

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

Explanation of Solution

The following unbalanced equation is given:

$\text{A}\text{s}{\text{I}}_3(s)\to \text{A}\text{s}(s)+{\text{I}}_2(s)$

The most complicated molecule is AsI₃. We start with Iodine. To balance Iodine, we place a coefficient ot 3 before I₂ and 2 before AsI₃

$2\text{A}\text{s}{\text{I}}_3(s)\to \text{A}\text{s}(s)+3{\text{I}}_2(s)$

To balance As, we place a coefficient of 2 before As.

$2\text{A}\text{s}{\text{I}}_3(s)\to 2\text{A}\text{s}(s)+3{\text{I}}_2(s)$

We now have a balanced equation.

(h)

Interpretation Introduction

Interpretation:

A balanced equation of the given reaction should be determined.

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 present before the compounds are changed but not the subscripts in the formulas.

Answer to Problem 44QAP

$2\text{C}\text{u}\text{S}{\text{O}}_4(aq)+4\text{K}\text{I}(s)\to 2\text{C}\text{u}\text{I}(s)+{\text{I}}_2(s)+2{\text{K}}_2\text{S}{\text{O}}_4(aq)$.

Explanation of Solution

The following unbalanced equation is given:

$\text{C}\text{u}\text{S}{\text{O}}_4(aq)+\text{K}\text{I}(s)\to \text{C}\text{u}\text{I}(s)+{\text{I}}_2(s)+{\text{K}}_2\text{S}{\text{O}}_4(aq)$

There is no clear choice for the most complicated molecule. We arbitrarily start with the elements of K₂ SO₄ To balance K, we place a coefficient of 2 before KI.

$\text{C}\text{u}\text{S}{\text{O}}_4(aq)+2\text{K}\text{I}(s)\to \text{C}\text{u}\text{I}(s)+{\text{I}}_2(s)+{\text{K}}_2\text{S}{\text{O}}_4(aq)$

We can balance the equation by giving I₂ a cofficient of ½.

$\text{C}\text{u}\text{S}{\text{O}}_4(aq)+\text{K}\text{I}(s)\to \text{C}\text{u}\text{I}(s)+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\text{I}}_2(s)+{\text{K}}_2\text{S}{\text{O}}_4(aq)$

The coffecient should be an integer, therefore we should multiply the equation by 2. We have got a balanced equation.

$2\text{C}\text{u}\text{S}{\text{O}}_4(aq)+4\text{K}\text{I}(s)\to 2\text{C}\text{u}\text{I}(s)+{\text{I}}_2(s)+2{\text{K}}_2\text{S}{\text{O}}_4(aq)$.