Chapter 11.1, Problem 4PP

(a)

Interpretation Introduction

Interpretation:

A chemical equation is given. It has to be balanced and possible mole ratio has to be determined.

$\text{ZnO}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{ZnCl}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Concept introduction:

According to law of conservation of mass it is not possible to form the mass or to destroy the mass. The mass always remains conserved. Thus, equation needs to be balanced.

Answer to Problem 4PP

The balanced chemical equation is shown below.

$\text{ZnO}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{ZnCl}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The mole ratio for given chemical equation is shown below.

$\begin{array}{l}\frac{\text{1}\text{mol ZnO}}{2\text{mol}\text{HCl}},\frac{\text{1}\text{mol ZnO}}{1\text{mol}{\text{ZnCl}}_2}\text{ and }\frac{\text{1}\text{mol ZnO}}{1\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{2\text{mol}\text{HCl}}{\text{1}\text{mol ZnO}},\frac{2\text{mol}\text{HCl}}{1\text{mol}{\text{ZnCl}}_2}\text{ and }\frac{2\text{mol}\text{HCl}}{1\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{1\text{mol}{\text{ZnCl}}_2}{\text{1}\text{mol ZnO}},\frac{1\text{mol}{\text{ZnCl}}_2}{2\text{mol}\text{HCl}}\text{ and }\frac{1\text{mol}{\text{ZnCl}}_2}{1\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{1\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{mol ZnO}},\frac{1\text{mol}{\text{H}}_{\text{2}}\text{O}}{2\text{mol}\text{HCl}}\text{ and }\frac{1\text{mol}{\text{H}}_{\text{2}}\text{O}}{1\text{mol}{\text{ZnCl}}_2}\end{array}$

Explanation of Solution

The given chemical equation is shown below.

$\text{ZnO}\left(s\right)+\text{HCl}\left(aq\right)\to {\text{ZnCl}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

A coefficient 2 is used and $\text{HCl}$ at left hand side is multiplied with it.

$\text{ZnO}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{ZnCl}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of zinc atom is one and oxygen atoms are also one on both the sides of the equation. The number of atoms of hydrogen and chlorine is two on both the sides. Hence, the above equation is a balanced chemical equation.

The mole ratio for given chemical equation is shown below.

$\begin{array}{l}\frac{\text{1}\text{mol ZnO}}{2\text{mol}\text{HCl}},\frac{\text{1}\text{mol ZnO}}{1\text{mol}{\text{ZnCl}}_2}\text{ and }\frac{\text{1}\text{mol ZnO}}{1\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{2\text{mol}\text{HCl}}{\text{1}\text{mol ZnO}},\frac{2\text{mol}\text{HCl}}{1\text{mol}{\text{ZnCl}}_2}\text{ and }\frac{2\text{mol}\text{HCl}}{1\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{1\text{mol}{\text{ZnCl}}_2}{\text{1}\text{mol ZnO}},\frac{1\text{mol}{\text{ZnCl}}_2}{2\text{mol}\text{HCl}}\text{ and }\frac{1\text{mol}{\text{ZnCl}}_2}{1\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{1\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{mol ZnO}},\frac{1\text{mol}{\text{H}}_{\text{2}}\text{O}}{2\text{mol}\text{HCl}}\text{ and }\frac{1\text{mol}{\text{H}}_{\text{2}}\text{O}}{1\text{mol}{\text{ZnCl}}_2}\end{array}$

(b)

Interpretation Introduction

Interpretation:

A chemical equation is given. It has to be balanced and possible mole ratio has to be determined.

Concept introduction:

According to law of conservation of mass it is not possible to form the mass or to destroy the mass. The mass always remains conserved. Thus, equation needs to be balanced.

Answer to Problem 4PP

The balanced chemical equation is shown below.

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}\left(g\right)+13{\text{O}}_{\text{2}}\left(g\right)\to 8{\text{CO}}_{\text{2}}\left(g\right)+10{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The mole ratio for given chemical equation is shown below.

$\begin{array}{l}\frac{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{13\text{mol}{\text{O}}_{\text{2}}},\frac{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{8\text{mol}{\text{CO}}_{\text{2}}}\text{and}\frac{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{10\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{13\text{mol}{\text{O}}_{\text{2}}}{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}},\frac{13\text{mol}{\text{O}}_{\text{2}}}{8\text{mol}{\text{CO}}_{\text{2}}}\text{and}\frac{13\text{mol}{\text{O}}_{\text{2}}}{10\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{8\text{mol}{\text{CO}}_{\text{2}}}{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}},\frac{8\text{mol}{\text{CO}}_{\text{2}}}{13\text{mol}{\text{O}}_{\text{2}}}\text{and}\frac{8\text{mol}{\text{CO}}_{\text{2}}}{10\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}},\frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{13\text{mol}{\text{O}}_{\text{2}}}\text{and}\frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{8\text{mol}{\text{CO}}_{\text{2}}}\end{array}$

Explanation of Solution

The given chemical equation is shown below.

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of carbon atoms is four at left hand side and one at right hand side of equation. Thus, carbon atoms are unbalanced. The number of hydrogen atoms is ten at the left hand side and two at the right hand side of the equation. Thus, hydrogen atoms are also unbalanced. The number of oxygen atoms is two at the left hand side and three at the right hand side of the equation. Thus, oxygen atoms are also unbalanced.

The coefficient of two is multiplied with butane, thirteen with oxygen, eight with carbon dioxide and ten with water for balancing the chemical equation.

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}\left(g\right)+13{\text{O}}_{\text{2}}\left(g\right)\to 8{\text{CO}}_{\text{2}}\left(g\right)+10{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of carbon atoms is eight at left hand side and at right hand side of equation. Thus, carbon atoms are balanced. The number of hydrogen atoms is twenty at the left hand side and at the right hand side of the equation. Thus, hydrogen atoms are also balanced. The number of oxygen atoms is twenty six at the left hand side and at the right hand side of the equation. Thus, oxygen atoms are also balanced.

The mole ratio for given chemical equation is shown below.

$\begin{array}{l}\frac{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{13\text{mol}{\text{O}}_{\text{2}}},\frac{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{8\text{mol}{\text{CO}}_{\text{2}}}\text{and}\frac{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{10\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{13\text{mol}{\text{O}}_{\text{2}}}{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}},\frac{13\text{mol}{\text{O}}_{\text{2}}}{8\text{mol}{\text{CO}}_{\text{2}}}\text{and}\frac{13\text{mol}{\text{O}}_{\text{2}}}{10\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{8\text{mol}{\text{CO}}_{\text{2}}}{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}},\frac{8\text{mol}{\text{CO}}_{\text{2}}}{13\text{mol}{\text{O}}_{\text{2}}}\text{and}\frac{8\text{mol}{\text{CO}}_{\text{2}}}{10\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{2}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}},\frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{13\text{mol}{\text{O}}_{\text{2}}}\text{and}\frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{8\text{mol}{\text{CO}}_{\text{2}}}\end{array}$