Chapter 6, Problem 40QAP

Balance each of the following chemical equations.

msp;  ${\text{Na}}_2{\text{SO}}_4\left(aq\right)+{\text{CaCl}}_2\left(aq\right)\to {\text{CaSO}}_4\left(s\right)+\text{NaCl}\left(aq\right)$

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

msp;  $\text{Ca}{\left(\text{OH}\right)}_2\left(aq\right)+\text{HCl}\left(aq\right)\to {\text{CaCl}}_2\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

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

msp;  $\text{NaOH}\left(s\right)+{\text{H}}_3{\text{PO}}_4\left(aq\right)\to {\text{Na}}_3{\text{PO}}_4\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

msp;  ${\text{Na}}_2{\text{O}}_2\left(s\right)+{\text{H}}_2\text{O}\left(l\right)\to \text{NaOH}\left(aq\right)+{\text{O}}_2\left(g\right)$

msp;  $\text{Si}\left(s\right)+{\text{S}}_8\left(s\right)\to {\text{Si}}_2{\text{S}}_4\left(s\right)$

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 40QAP

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

Explanation of Solution

The provided reaction is:

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

The most complicated molecule is not clear, so we start arbitrarily. If we place a coefficient of 2NaCl, we get a balanced equation.

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

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 40QAP

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

Explanation of Solution

The given unbalanced equation is:

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

We start with placing 3 before Fe and 4 before H₂ O and H₂ We get the balanced chemical reaction as follows:

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

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 40QAP

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

Explanation of Solution

The given unbalanced equation is:

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

We start by balancing HCl. If we place a coefficient 2 before HCl

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

If we place 2 before H₂ O, we get a balanced chemical reaction as follows:

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

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 40QAP

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

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.

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

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 40QAP

$3\text{N}\text{a}\text{O}\text{H}(s)+{\text{H}}_3\text{P}{\text{O}}_4(aq)\to \text{N}{\text{a}}_3\text{P}{\text{O}}_4(aq)+3{\text{H}}_2\text{O}(l)$.

Explanation of Solution

The provided equation is:

$\text{N}\text{a}\text{O}\text{H}(s)+{\text{H}}_3\text{P}{\text{O}}_4(aq)\to \text{N}{\text{a}}_3\text{P}{\text{O}}_4(aq)+{\text{H}}_2\text{O}(l)$

We start with NaOH. To balance Na. If we place a coefficient of 3 before NaOH

$3\text{N}\text{a}\text{O}\text{H}(s)+{\text{H}}_3\text{P}{\text{O}}_4(aq)\to \text{N}{\text{a}}_3\text{P}{\text{O}}_4(aq)+{\text{H}}_2\text{O}(l)$

To balance oxygen, we place 3 before H₂ O. We get the balanced chemical reaction as follows:

$3\text{N}\text{a}\text{O}\text{H}(s)+{\text{H}}_3\text{P}{\text{O}}_4(aq)\to \text{N}{\text{a}}_3\text{P}{\text{O}}_4(aq)+3{\text{H}}_2\text{O}(l)$.

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 40QAP

$2\text{N}\text{a}\text{N}{\text{O}}_3(s)\to 2\text{N}\text{a}\text{N}{\text{O}}_2(s)+{\text{O}}_2(g)$.

Explanation of Solution

The provided reaction is:

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

To balance oxygen, we can start by placing 2 before NaNO₂ and NaNO₃. Now, we already have a balanced equation.

$2\text{N}\text{a}\text{N}{\text{O}}_3(s)\to 2\text{N}\text{a}\text{N}{\text{O}}_2(s)+{\text{O}}_2(g)$.

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 40QAP

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

Explanation of Solution

The provided reaction is:

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

We start with balancing $\text{N}{\text{a}}_2{\text{O}}_2$. We place a coefficient 2 before $\text{N}{\text{a}}_2{\text{O}}_2$

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

Now we can place 4 before NaOH and 2 before H₂ O. We have a balanced equation

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

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 40QAP

$4\text{S}\text{i}(s)+{\text{S}}_8(s)\to 2\text{S}{\text{i}}_2{\text{S}}_4(s)$.

Explanation of Solution

The provided reaction is:

$\text{S}\text{i}(s)+{\text{S}}_8(s)\to \text{S}{\text{i}}_2{\text{S}}_4(s)$

$\text{S}{\text{i}}_2{\text{S}}_4$ is the most complicated element. It contains the most number of elements and we will start with the elements of $\text{S}{\text{i}}_2{\text{S}}_4$. If we place a coefficient of 2 before $\text{S}{\text{i}}_2{\text{S}}_4$

$\text{S}\text{i}(s)+{\text{S}}_8(s)\to 2\text{S}{\text{i}}_2{\text{S}}_4(s)$

Now we place 4 before Si and we have a balanced equation.

$4\text{S}\text{i}(s)+{\text{S}}_8(s)\to 2\text{S}{\text{i}}_2{\text{S}}_4(s)$.