Chapter 6, Problem 39QAP

Balance each of the following chemical equations.

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

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

msp;  $\text{NaOH}\left(aq\right)+{\text{HClO}}_4\left(aq\right)\to {\text{NaCIO}}_4\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

msp;  $\text{Mg}\left(s\right)+{\text{Mn}}_2{\text{O}}_3\left(s\right)\to \text{MgO}\left(s\right)+\text{Mn}\left(s\right)$

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

msp;  ${\text{NO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{O}}_2\left(g\right)\to {\text{HNO}}_3\left(aq\right)$

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

msp;  ${\text{NH}}_3\left(g\right)+{\text{O}}_2\left(g\right)\to \text{NO}\left(g\right)+{\text{H}}_2\text{O}\left(l\right)$

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 39QAP

${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{(aq)+BaCl}}_{\text{2}}\text{(aq)}\to {\text{BaSO}}_{\text{4}}\text{(s)+2KCl(aq)}$.

Explanation of Solution

The provided reaction is:

${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{(aq)+BaCl}}_{\text{2}}\text{(aq)}\to {\text{BaSO}}_{\text{4}}\text{(s)+KCl(aq)}$

If we place a coefficient of 2 before KCl in the product side as there is already two potassium on the reactant side, then we have a balanced equation. All the other atoms are balanced and need not require any coefficients. The balancing equation is given as follows,

${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{(aq)+BaCl}}_{\text{2}}\text{(aq)}\to {\text{BaSO}}_{\text{4}}\text{(s)+2KCl(aq)}$.

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 39QAP

${\text{Fe(s)+H}}_{\text{2}}\text{O(s)}\to {\text{FeO(s)+H}}_{\text{2}}\text{(g)}$.

Explanation of Solution

The given unbalanced equation is:

${\text{Fe(s)+H}}_{\text{2}}\text{O(s)}\to {\text{FeO(s)+H}}_{\text{2}}\text{(g)}$

All the atoms are balanced such as two hydrogen, one oxygen and one iron in the given equation and so it requires no coefficients on either side. Thus, the balanced chemical reaction is as follows:

${\text{Fe(s)+H}}_{\text{2}}\text{O(s)}\to {\text{FeO(s)+H}}_{\text{2}}\text{(g)}$.

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 39QAP

${\text{NaOH(aq)+HClO}}_{\text{4}}\text{(aq)}\to {\text{NaClO}}_{\text{4}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}$.

Explanation of Solution

The given unbalanced equation is:

${\text{NaOH(aq)+HClO}}_{\text{4}}\text{(aq)}\to {\text{NaClO}}_{\text{4}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}$

All the atoms are balanced in the given equation with one sodium, two hydrogen and five oxygen atoms and so no balance is required on either side of the equation. Thus, the balanced chemical reaction is as follows:

${\text{NaOH(aq)+HClO}}_{\text{4}}\text{(aq)}\to {\text{NaClO}}_{\text{4}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}$.

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 39QAP

${\text{3Mg(s)+Mn}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\to \text{3MgO(s)+2Mn(s)}$.

Explanation of Solution

The provided equation is:

${\text{Mg(s)+Mn}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\to \text{MgO(s)+Mn(s)}$

Coefficient 3 should be inserted before MgO on the product side, coefficient 2 before the manganese Mn on the product side, and coefficient 3 before magnesium Mg on the reactant side to balance the equation. Thus, the balanced chemical reaction is as follows:

${\text{3Mg(s)+Mn}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\to \text{3MgO(s)+2Mn(s)}$.

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 39QAP

${\text{2KOH(s)+KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{(aq)}\to {\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}$.

Explanation of Solution

The provided equation is:

${\text{KOH(s)+KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{(aq)}\to {\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}$

The most complicated molecule is KH₂ PO₄ because it contains the greatestnumber of elements. If we place a coefficient of 2 before KOH on the reactant side and a coefficient of 2 before H₂ O on the product side, it gives two atoms of potassium, four atoms of hydrogen, six atoms of oxygen and one atom of phosphorous, the balanced chemical reaction is as follows:

${\text{2KOH(s)+KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{(aq)}\to {\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}$.

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 39QAP

${\text{4NO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}{\text{O(l)+O}}_{\text{2}}\text{(g)}\to {\text{4HNO}}_{\text{3}}\text{(aq)}$.

Explanation of Solution

The provided reaction is:

${\text{NO}}_{\text{2}}{\text{(g)+H}}_{\text{2}}{\text{O(l)+O}}_{\text{2}}\text{(g)}\to {\text{HNO}}_{\text{3}}\text{(aq)}$

If we place a coefficient of 2 before H₂ O anda coefficient of 4 before HNO₃ we get

${\text{NO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}{\text{O(l)+O}}_{\text{2}}\text{(g)}\to {\text{4HNO}}_{\text{3}}\text{(aq)}$

And we place a coefficient of 4 before NO₂.

Now, we have a balanced equation.

${\text{4NO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}{\text{O(l)+O}}_{\text{2}}\text{(g)}\to {\text{4HNO}}_{\text{3}}\text{(aq)}$.

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 39QAP

${\text{2BaO}}_{\text{2}}{\text{(s)+2H}}_{\text{2}}\text{O(l)}\to {\text{2Ba(OH)}}_{\text{2}}{\text{(aq)+O}}_{\text{2}}\text{(g)}$.

Explanation of Solution

The provided reaction is:

${\text{BaO}}_{\text{2}}{\text{(s)+H}}_{\text{2}}\text{O(l)}\to {\text{Ba(OH)}}_{\text{2}}{\text{(aq)+O}}_{\text{2}}\text{(g)}$

If we place a coefficient of 2 before BaO₂, 2 before H₂ O on the product side and 2 before Ba(OH)₂ on the reactant side that the equation will be balanced. And we have a balanced equation.

${\text{2BaO}}_{\text{2}}{\text{(s)+2H}}_{\text{2}}\text{O(l)}\to {\text{2Ba(OH)}}_{\text{2}}{\text{(aq)+O}}_{\text{2}}\text{(g)}$.

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 39QAP

${\text{4NH}}_{\text{3}}{\text{(g)+5O}}_{\text{2}}\text{(g)}\to {\text{4NO(g)+6H}}_{\text{2}}\text{O(g)}$.

Explanation of Solution

The provided reaction is:

${\text{NH}}_{\text{3}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{NO(g)+H}}_{\text{2}}\text{O(g)}$

We should start by balancing H by placing a coefficient of 3 before H₂ O and 2 before NH₃.

${\text{2NH}}_{\text{3}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{NO(g)+3H}}_{\text{2}}\text{O(g)}$

We can balance NO by adding a coefficient of 2 before NO,

${\text{2NH}}_{\text{3}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{2NO(g)+3H}}_{\text{2}}\text{O(g)}$

Now to balance oxygen, we should place 5 before O₂, 6 before H₂ O, 4 before NO and 4 before NH₃. Now, we have a balanced equation.

${\text{4NH}}_{\text{3}}{\text{(g)+5O}}_{\text{2}}\text{(g)}\to {\text{4NO(g)+6H}}_{\text{2}}\text{O(g)}$.