Chapter 9, Problem 58E

Interpretation Introduction

(a)

Interpretation:

The table is to be completed according to the balanced chemical equation.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in a reaction which determines the moles of the reactants and products in the reaction. The ratio of moles is termed as mole ratio. In stoichiometry problems, the reactant that controls the amount of the product formed is known as the limiting reactant.

Answer to Problem 58E

The table is completed as shown below.

Experiment $\text{mol}\text{Co}$ $\text{mol}\text{S}$ $\text{mol}{\text{Co}}_2{\text{S}}_3$

(a) before reaction: $1.00$ $1.00$ $0.00$

after reaction: $0.67$ $0.00$ $0.33$

Explanation of Solution

The table to be completed is given below.

Experiment $\text{mol}\text{Co}$ $\text{mol}\text{S}$ $\text{mol}{\text{Co}}_2{\text{S}}_3$

(a) before reaction: $1.00$ $1.00$ $0.00$

after reaction:

The balanced equation is given below.

$\text{2Co}\left(s\right)\text{+3S}\left(s\right)\stackrel{\Delta }{\to }{\text{Co}}_2{\text{S}}_3\left(s\right)$

In the reaction, $2$ moles of $\text{Co}$ produce $1$ mole of ${\text{Co}}_2{\text{S}}_3$.

Therefore, the mole ratio is given below.

$\frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{2\text{mol}\text{Co}}$ and $\frac{2\text{mol}\text{Co}}{1\text{mol}{\text{Co}}_2{\text{S}}_3}$

The mole ratio to obtain moles of ${\text{Co}}_2{\text{S}}_3$ from moles of $\text{Co}$ is given below.

$\frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{2\text{mol}\text{Co}}$

The formula to calculate the moles of ${\text{Co}}_2{\text{S}}_3$ from moles of $\text{Co}$ is given below.

$\text{Moles}\text{of}{\text{Co}}_2{\text{S}}_3=\left(\begin{array}{l}\text{Given}\text{moles}\text{of}\text{Co}\\ \times \text{Mole}\text{ratio}\text{to}\text{obtain}\text{moles}\text{of}{\text{Co}}_2{\text{S}}_3\text{from}\text{Co}\end{array}\right)$ …(1)

The moles of $\text{Co}$ is $\text{1}\text{.00}\text{mol}$.

Substitute the value of moles of $\text{Co}$ and mole ratio in equation (1).

$\begin{array}{rll}\text{Moles}\text{of}{\text{Co}}_2{\text{S}}_3& =& 1.\text{00}\text{mol}\text{Co}\times \frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{2\text{mol}\text{Co}}\\ & =& 0.5\text{mol}\end{array}$

In the reaction, $3$ moles of $\text{S}$ produce $1$ mole of ${\text{Co}}_2{\text{S}}_3$.

Therefore, the mole ratio is given below.

$\frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{3\text{mol}\text{S}}$ and $\frac{3\text{mol}\text{S}}{1\text{mol}{\text{Co}}_2{\text{S}}_3}$

The mole ratio to obtain moles of ${\text{Co}}_2{\text{S}}_3$ from moles of $\text{S}$ is given below.

$\frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{3\text{mol}\text{S}}$

The formula to calculate the moles of ${\text{Co}}_2{\text{S}}_3$ from moles of $\text{S}$ is given below.

$\text{Moles}\text{of}{\text{Co}}_2{\text{S}}_3=\left(\begin{array}{l}\text{Given}\text{moles}\text{of}\text{S}\\ \times \text{Mole}\text{ratio}\text{to}\text{obtain}\text{moles}\text{of}{\text{Co}}_2{\text{S}}_3\text{from}\text{S}\end{array}\right)$ …(2)

The moles of $\text{S}$ is $\text{1}\text{.00}\text{mol}$.

Substitute the value of moles of $\text{S}$ and mole ratio in equation (2).

$\begin{array}{rll}\text{Moles}\text{of}{\text{Co}}_2{\text{S}}_3& =& 1.\text{00}\text{mol}\text{S}\times \frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{3\text{mol}\text{S}}\\ & =& 0.33\text{mol}\end{array}$

Since, $\text{1}\text{.00}\text{mol}$ of $\text{S}$ produces lesser amount of ${\text{Co}}_2{\text{S}}_3$, $\text{S}$ is the limiting reactant. The moles of ${\text{Co}}_2{\text{S}}_3$ that are produced from the reaction are $0.33\text{mol}$.

After the reaction, the moles of $\text{S}$ will be completely used.

The moles of $\text{Co}$ left are calculated below.

$\begin{array}{rll}\text{Moles}\text{of}\text{Co}& =& 1\text{mol}-\text{0}\text{.33}\text{mol}\\ & =& \text{0}\text{.67}\text{mol}\end{array}$

The table is therefore completed as shown below.

Experiment $\text{mol}\text{Co}$ $\text{mol}\text{S}$ $\text{mol}{\text{Co}}_2{\text{S}}_3$

(a) before reaction: $1.00$ $1.00$ $0.00$

after reaction: $0.67$ $0.00$ $0.33$

Conclusion

The table is completed has been rightfully completed.

Interpretation Introduction

(b)

Interpretation:

The table is to be completed according to the balanced chemical equation.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in a reaction which determines the moles of the reactants and products in the reaction. The ratio of moles is termed as mole ratio. In stoichiometry problems, the reactant that controls the amount of the product formed is known as the limiting reactant.

Answer to Problem 58E

The table is completed as shown below.

Experiment $\text{mol}\text{Co}$ $\text{mol}\text{S}$ $\text{mol}{\text{Co}}_2{\text{S}}_3$

(b) before reaction: $2.00$ $3.00$ $0.00$

after reaction: $0.00$ $1.00$ $2.00$

Explanation of Solution

The table to be completed is given below.

Experiment $\text{mol}\text{Co}$ $\text{mol}\text{S}$ $\text{mol}{\text{Co}}_2{\text{S}}_3$

(b) before reaction: $2.00$ $3.00$ $0.00$

after reaction:

The balanced equation is given below.

$\text{2Co}\left(s\right)\text{+3S}\left(s\right)\stackrel{\Delta }{\to }{\text{Co}}_2{\text{S}}_3\left(s\right)$

In the reaction, $2$ moles of $\text{Co}$ produce $1$ mole of ${\text{Co}}_2{\text{S}}_3$.

Therefore, the mole ratio is given below.

$\frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{2\text{mol}\text{Co}}$ and $\frac{2\text{mol}\text{Co}}{1\text{mol}{\text{Co}}_2{\text{S}}_3}$

The mole ratio to obtain moles of ${\text{Co}}_2{\text{S}}_3$ from moles of $\text{Co}$ is given below.

$\frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{2\text{mol}\text{Co}}$

The formula to calculate the moles of ${\text{Co}}_2{\text{S}}_3$ from moles of $\text{Co}$ is given below.

$\text{Moles}\text{of}{\text{Co}}_2{\text{S}}_3=\left(\begin{array}{l}\text{Given}\text{moles}\text{of}\text{Co}\\ \times \text{Mole}\text{ratio}\text{to}\text{obtain}\text{moles}\text{of}{\text{Co}}_2{\text{S}}_3\text{from}\text{Co}\end{array}\right)$ …(1)

The moles of $\text{Co}$ is $\text{2}\text{.00}\text{mol}$.

Substitute the value of moles of $\text{Co}$ and mole ratio in equation (1).

$\begin{array}{rll}\text{Moles}\text{of}{\text{Co}}_2{\text{S}}_3& =& 2.\text{00}\text{mol}\text{Co}\times \frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{2\text{mol}\text{Co}}\\ & =& 1.00\text{mol}\end{array}$

In the reaction, $3$ moles of $\text{S}$ produce $1$ mole of ${\text{Co}}_2{\text{S}}_3$.

Therefore, the mole ratio is given below.

$\frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{3\text{mol}\text{S}}$ and $\frac{3\text{mol}\text{S}}{1\text{mol}{\text{Co}}_2{\text{S}}_3}$

The mole ratio to obtain moles of ${\text{Co}}_2{\text{S}}_3$ from moles of $\text{S}$ is given below.

$\frac{1\text{mol}\text{CoS}}{3\text{mol}\text{S}}$

The formula to calculate the moles of ${\text{Co}}_2{\text{S}}_3$ from moles of $\text{S}$ is given below.

$\text{Moles}\text{of}{\text{Co}}_2{\text{S}}_3=\left(\begin{array}{l}\text{Given}\text{moles}\text{of}\text{S}\\ \times \text{Mole}\text{ratio}\text{to}\text{obtain}\text{moles}\text{of}{\text{Co}}_2{\text{S}}_3\text{from}\text{S}\end{array}\right)$ …(2)

The moles of $\text{S}$ is $\text{3}\text{.00}\text{mol}$.

Substitute the value of moles of $\text{S}$ and mole ratio in equation (2).

$\begin{array}{rll}\text{Moles}\text{of}{\text{Co}}_2{\text{S}}_3& =& 3.\text{00}\text{mol}\text{S}\times \frac{1\text{mol}{\text{Co}}_2{\text{S}}_3}{3\text{mol}\text{S}}\\ & =& 1\text{mol}\end{array}$

Since, both $\text{Co}$ and $\text{S}$ produce $1.00\text{mol}$ of ${\text{Co}}_2{\text{S}}_3$, they both are limiting reactants. They both are completely used in the reaction. The moles of ${\text{Co}}_2{\text{S}}_3$ that are produced from the reaction is $1.00\text{mol}$.

The table is therefore completed as shown below.

Experiment $\text{mol}\text{Co}$ $\text{mol}\text{S}$ $\text{mol}{\text{Co}}_2{\text{S}}_3$

(b) before reaction: $2.00$ $3.00$ $0.00$

after reaction: $0.00$ $0.00$ $1.00$

Conclusion

The table is completed has been rightfully completed.