Chapter 3, Problem 48QAP

Balance the following equations:

(a) ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(\text{g}\right)+{\text{H}}_2\text{O}\left(l\right)$

(b) ${\text{XeF}}_4\left(g\right)+{\text{H}}_2\text{O}\left(l\right)\to \text{Xe}\left(g\right)+{\text{O}}_2\left(g\right)+\text{HF}\left(g\right)$

(c) $\text{NaCl}\left(s\right)+{\text{H}}_2\text{O}\left(g\right)+{\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{Na}}_2{\text{SO}}_4\left(s\right)+{\text{Na}}_2{\text{SO}}_4\left(s\right)+\text{HCl}\left(g\right)$

Interpretation Introduction

(a)

Interpretation:

The given chemical equation should be balanced.

Concept introduction:

The chemical reaction is said to be balanced if there are same number of constituent atoms are present on both sides of the reaction arrow. A balanced chemical reaction satisfies the law of conservation of mass.

Answer to Problem 48QAP

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)+6{\text{O}}_2\left(g\right)\to 6{\text{CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given chemical reaction is as follows:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of carbon atoms on left hand side is 6 and that on the right hand side is 1 thus, give coefficient 6 to ${\text{CO}}_{\text{2}}$ to balance the number of carbon atoms.

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)+{\text{O}}_2\left(g\right)\to 6{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number hydrogen atoms on left hand side is 12 and that on the right hand side is 2 thus, give coefficient 6 to ${\text{H}}_{\text{2}}\text{O}$ to balance the number of hydrogen atoms.

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)+{\text{O}}_2\left(g\right)\to 6{\text{CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Now, number of oxygen atoms on left hand side of the reaction arrow is 8 and that on the right hand side is 18 thus, give coefficient 6 to ${\text{O}}_{\text{2}}$ on the left hand side to balance the number of oxygen atoms.

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)+6{\text{O}}_2\left(g\right)\to 6{\text{CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Thus, the balanced chemical reaction will be:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)+6{\text{O}}_2\left(g\right)\to 6{\text{CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Interpretation Introduction

(b)

Interpretation:

The given chemical equation should be balanced.

Concept introduction:

The chemical reaction is said to be balanced if there are same number of constituent atoms are present on both sides of the reaction arrow. A balanced chemical reaction satisfies the law of conservation of mass.

Answer to Problem 48QAP

${\text{XeF}}_{\text{4}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Xe}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)+4\text{HF}\left(g\right)$

Explanation of Solution

The given chemical reaction is as follows:

${\text{XeF}}_{\text{4}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Xe}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)+\text{HF}\left(g\right)$

Give coefficient 4 to HF to balance the number of F atoms.

${\text{XeF}}_{\text{4}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Xe}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)+4\text{HF}\left(g\right)$

Now, give coefficient 2 to ${\text{H}}_{\text{2}}\text{O}$ on the reactant side to balance the number of H atoms.

${\text{XeF}}_{\text{4}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Xe}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)+4\text{HF}\left(g\right)$

The number of oxygen atoms on left and right hand side of the reaction arrow is same. Thus, the balanced chemical reaction will be:

${\text{XeF}}_{\text{4}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Xe}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)+4\text{HF}\left(g\right)$

Interpretation Introduction

(c)

Interpretation:

The given chemical equation should be balanced.

Concept introduction:

The chemical reaction is said to be balanced if there are same number of constituent atoms are present on both sides of the reaction arrow. A balanced chemical reaction satisfies the law of conservation of mass.

Answer to Problem 48QAP

$\text{4NaCl}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)+2{\text{SO}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)+4\text{HCl}\left(g\right)$

Explanation of Solution

The given reaction is as follows:

$\text{NaCl}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{SO}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)+\text{HCl}\left(g\right)$

The number of sulfur atom is already balanced. To balance the number of sodium atom, give coefficient 2 to NaCl thus,

$\text{2NaCl}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{SO}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)+\text{HCl}\left(g\right)$

Now, to balance the number of chlorine atoms, give coefficient 2 to HCl thus,

$\text{2NaCl}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{SO}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)+2\text{HCl}\left(g\right)$

The number of oxygen atom on left and side is 5 and that on right hand side is 4 thus, give coefficient ½ to ${\text{O}}_{\text{2}}$ to balance the number of oxygen atom.

$\text{2NaCl}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{SO}}_{\text{2}}\left(g\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)+2\text{HCl}\left(g\right)$

Or,

$\text{4NaCl}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)+2{\text{SO}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)+4\text{HCl}\left(g\right)$

Thus, the balanced chemical reaction is as follows:

$\text{4NaCl}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)+2{\text{SO}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)+4\text{HCl}\left(g\right)$