Chapter 3, Problem 47QAP

Balance the following equations:

(a) ${\text{CaC}}_2\left(s\right)+{\text{H}}_2\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_2\left(s\right)+{\text{C}}_2{\text{H}}_2\left(g\right)$

(b) ${\left({\text{NH}}_4\right)}_2{\text{Cr}}_2{\text{O}}_7\left(s\right)\to {\text{Cr}}_2{\text{O}}_3\left(s\right)+{\text{N}}_2\left(g\right)+{\text{H}}_2\text{O}\left(f\right)$

(c) ${\text{CH}}_3{\text{NH}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{N}}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

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 47QAP

${\text{CaC}}_{\text{2}}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$

Explanation of Solution

The given chemical reaction is as follows:

${\text{CaC}}_{\text{2}}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$

The number of Ca and C atom are already balanced in the given reaction. The number of oxygen atoms on the left side of the reaction arrow is 1 and that on the right hand side of the reaction arrow is 2.

Similarly, the number of hydrogen atoms on the left side of the reaction arrow is 2 and that of the oxygen atoms is 4.

Thus, give coefficient 2 to ${\text{H}}_{\text{2}}\text{O}$ on the left hand side of the reaction arrow to balance the number of hydrogen and oxygen atoms.

${\text{CaC}}_{\text{2}}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$

Thus, the balanced chemical reaction will be:

${\text{CaC}}_{\text{2}}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$

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 47QAP

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(s\right)\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{N}}_{\text{2}}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The given chemical reaction is as follows:

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(s\right)\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{N}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The number of N and Cr atom is already balanced. The number of hydrogen atoms on the left side of the reaction arrow is 8 and that on the right hand side is 2 thus, give coefficient 4 to ${\text{H}}_{\text{2}}\text{O}$ to balance the number of oxygen atoms.

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(s\right)\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{N}}_{\text{2}}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Now, number of oxygen atoms on left hand side is 7 and that on the right hand side is also 7. Therefore, the balanced chemical reaction will be:

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(s\right)\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{N}}_{\text{2}}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(g\right)$

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 47QAP

${\text{4CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+9{\text{O}}_2\left(g\right)\to 4{\text{CO}}_{\text{2}}\left(g\right)+2{\text{N}}_{\text{2}}\left(g\right)+10{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The given chemical reaction is as follows:

${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)+{\text{N}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The number of nitrogen atom on left hand side is 1 and that on the right hand side is 2, give coefficient 2 to ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$.

${\text{2CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)+{\text{N}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

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

${\text{2CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{CO}}_{\text{2}}\left(g\right)+{\text{N}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The number of hydrogen atom on the left hand side is 10 and that on the right hand side is 2 thus, give coefficient 5 to ${\text{H}}_{\text{2}}\text{O}$ to balance the number of oxygen atom.

${\text{2CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{CO}}_{\text{2}}\left(g\right)+{\text{N}}_{\text{2}}\left(g\right)+5{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The number of oxygen atom on left hand side is 2 and that on the right hand side is 9. To balance the oxygen atoms, give coefficient 9/2 to oxygen gas on the left hand side.

${\text{2CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+\frac{9}{2}{\text{O}}_2\left(g\right)\to 2{\text{CO}}_{\text{2}}\left(g\right)+{\text{N}}_{\text{2}}\left(g\right)+5{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Or,

${\text{4CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+9{\text{O}}_2\left(g\right)\to 4{\text{CO}}_{\text{2}}\left(g\right)+2{\text{N}}_{\text{2}}\left(g\right)+10{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Thus, the balanced chemical reaction is as follows:

${\text{4CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+9{\text{O}}_2\left(g\right)\to 4{\text{CO}}_{\text{2}}\left(g\right)+2{\text{N}}_{\text{2}}\left(g\right)+10{\text{H}}_{\text{2}}\text{O}\left(g\right)$