Chapter 5, Problem 5.99P

Interpretation Introduction

(a)

Interpretation:

The number of moles and number of grams of ${\text{NO}}_2$ formed in ${\text{2NO + O}}_2\to 2{\text{ NO}}_2$ given that 1.0 mol of NO and 1.0 mol of ${\text{O}}_2$ are involved should be determined.

Concept Introduction:

The formula used to determine the number of moles is:

  $\text{number of moles=}\frac{\text{given mass}}{\text{MolarMass}}$

Answer to Problem 5.99P

The number of moles of the product is 1 mole and the number of grams of the product is 46g.

Explanation of Solution

Given that 1.0 mol of NO and 1.0 mol of ${\text{O}}_2$ are involved in the reaction of ${\text{2NO + O}}_2\to 2{\text{ NO}}_2$

  $\begin{array}{c}2{\text{ mole of NO requires 1 mole of O}}_2\\ {\text{Therefore, 1 mole of NO requires 'x' moles of O}}_2\\ \text{i}\text{.e; 2 mole =1 mole}\\ \text{1 mole = }\frac{\text{1 x 1 }}{\text{2}}\\ \text{ = }\frac{\text{1}}{\text{2}}\text{mole}\\ \text{= 0}\text{.5 mole}\end{array}$

It can be observed that 1 mole needs only 0.5 mole of oxygen to complete the reaction. But actually, 1 mole is said to be involved indicating that there is an excess of oxygen. Hence, the limiting reactant is NO.

To compute the number of moles, the following steps are taken

Consider the equation ${\text{2NO + O}}_2\to 2{\text{ NO}}_2$ wherein

  $\begin{array}{l}{\text{2 moles of NO gives 2 moles of NO}}_2\\ \text{Therefore, 1mole will produce 1 mole of product}\text{.}\end{array}$

Hence, the number of moles of the product is 1 mole.

To compute the number of grams, it is known that the molar mass of ${\text{NO}}_2$ is 46.01g/mol

Hence, 1 mole = 46.01 g

Therefore, number of grams of ${\text{NO}}_2$ is 46 g.

Interpretation Introduction

(b)

Interpretation:

The number of moles and number of grams of ${\text{NO}}_2$ formed in ${\text{2NO + O}}_2\to 2{\text{ NO}}_2$ given that 2.0 mol of NO and 0.5 mol of ${\text{O}}_2$ are involved should be determined.

Concept Introduction:

The formula used to determine the number of moles is:

  $\text{number of moles=}\frac{\text{given mass}}{\text{MolarMass}}$

Answer to Problem 5.99P

The number of moles of the product is 1 mole and the number of grams of the product is 46 g.

Explanation of Solution

Given that 2.0 mol of NO and 0.5 mol of ${\text{O}}_2$ are involved in the reaction of ${\text{2NO + O}}_2\to 2{\text{ NO}}_2$

For complete reaction, 2 mole of NO requires 1 mole of ${\text{O}}_2$

But actual condition indicates that 0.5 mole of oxygen is used indicating that there is less moles of oxygen. This indicates that the limiting reactant is ${\text{O}}_2$

To compute the number of moles, the following steps are taken

Consider the equation ${\text{2NO + O}}_2\to 2{\text{ NO}}_2$ wherein

2 moles of NO and 1 mole of oxygen react to give 2 moles of product.

Ratio of reactant is $\frac{\text{2 moles of NO}}{{\text{1mole of O}}_2}$

But for 0.5 mole of ${\text{O}}_2$

  $\begin{array}{c}\frac{\text{2 moles of NO}}{{\text{1mole of O}}_2}=\frac{x}{\text{0}\text{.5moles}}\\ =2\times \text{ 0}\text{.5}\\ \text{=1}\text{.0 mole of NO}\end{array}$

Hence, the number of moles of NO is 1 mole. Therefore, only 1 mole of product is formed

To compute the number of grams, it is known that the molar mass of ${\text{NO}}_2$ is 46.01g/mol

Hence, 1 mole = 46.01 g

Therefore, the number of grams of ${\text{NO}}_2$ is 46 g

Interpretation Introduction

(c)

Interpretation:

The number of moles and number of grams of ${\text{NO}}_2$ formed in ${\text{2NO + O}}_2\to 2{\text{ NO}}_2$ given that 10 g of NO and 10 g of ${\text{O}}_2$ are involved should be determined.

Concept Introduction:

The formula used to determine the number of moles is:

  $\text{number of moles=}\frac{\text{given mass}}{\text{MolarMass}}$

Answer to Problem 5.99P

The number of moles of the product is 0.333 mole and the number of grams of the product is 15.3 g.

Explanation of Solution

Given

the mass of nitric oxide =10g

Molar mass of Nitric oxide =30.01g/mol

The number of moles of NO is computed as

  $\begin{array}{c}\text{1 mole of NO=30}\text{.01 g}\\ x\text{ mole of NO=10g}\\ \text{30}\text{.0 }x\text{=10}\\ x\text{=}\frac{10}{30}=0.333\text{ mole}\end{array}$

The mass of oxygen =10 g

Molar mass of oxygen =32.0 g/mol

The number of moles of Oxygen is computed as

  $\begin{array}{c}{\text{1 mole of O}}_2\text{=32}\text{.0 g}\\ x{\text{ mole of O}}_2\text{=10g}\\ \text{32}\text{.0 }x\text{=10}\\ x\text{=}\frac{10}{32}=0.3125\text{ mole}\\ \approx \text{0}\text{.313 mole}\end{array}$

Hence, no. of moles of NO is 0.333 mole

No. of moles of ${\text{NO}}_2$ is 0.313 mole

For 1 mole of NO to react with oxygen, the number of moles of oxygen necessary is computed as

  $\begin{array}{c}{\text{2 mole of NO=1 mole of O}}_2\\ 0.333\text{ mole of NO=}\frac{0.333}{2}\\ \text{=0}\text{.1665 mole}\\ \approx \text{0}\text{.167 mole}\end{array}$

But previously it is noted that 0.333 mole of NO reacts with 0.313 mole of oxygen instead of 0.167 mole. Hence, it is clearly seen that there is an excess of oxygen involved indicating that NO is the limiting reactant.

To calculate the number of moles of products

  $\begin{array}{c}{\text{2NO + O}}_2\to 2{\text{ NO}}_2\\ {\text{2 mole of NO=2 mole of NO}}_2\\ 0.333\text{ mole of NO=0}{\text{.333 mole of NO}}_2\end{array}$

Therefore, 0.333 mole of products is formed

To calculate the number of grams of the product

Molar mass of ${\text{NO}}_2$ = 46.01 g/mol

  $\begin{array}{c}{\text{1 mole of NO}}_2=46.01g\\ 0.33{\text{3 mole of NO}}_2=0.333\text{ x 46}\text{.01}\\ \text{=15}\text{.321g}\\ \approx \text{15}\text{.3g}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The number of moles and number of grams of ${\text{NO}}_2$ formed in ${\text{2NO + O}}_2\to 2{\text{ NO}}_2$ given that 28g of NO and 16 g of ${\text{O}}_2$ are involved should be determined.

Concept Introduction:

The formula used to determine the number of moles is:

  $\text{number of moles=}\frac{\text{given mass}}{\text{MolarMass}}$

Answer to Problem 5.99P

The number of moles of the product is 0.933 mole and the number of grams of the product is 42.93g.

Explanation of Solution

Given

the mass of nitric oxide =10g

Molar mass of Nitric oxide =30.01g/mol

The number of moles of NO is computed as

  $\begin{array}{c}\text{1 mole of NO=30}\text{.01 g}\\ x\text{ mole of NO=28g}\\ \text{30}\text{.0 }x\text{=28}\\ x\text{=}\frac{28}{30}=0.933\text{ mole}\end{array}$

the mass of oxygen =10g

Molar mass of oxygen =32.0 g/mol

The number of moles of Oxygen is computed as

  $\begin{array}{c}{\text{1 mole of O}}_2\text{=32}\text{.0 g}\\ x{\text{ mole of O}}_2\text{=16g}\\ \text{32}\text{.0 }x\text{=16}\\ x\text{=}\frac{16}{32}=0.5\text{ mole}\end{array}$

Hence, no. of moles of NO is 0.933 mole

No. of moles of ${\text{NO}}_2$ is 0.5 mole

For 1 mole of NO to react with oxygen, the number of moles of oxygen necessary is computed as

  $\begin{array}{c}{\text{2 mole of NO=1 mole of O}}_2\\ 0.933\text{ mole of NO=}\frac{0.933}{2}\\ \text{=0}\text{.4665 mole}\\ \approx \text{0}\text{.467 mole}\end{array}$

But previously it is noted that 0.933 mole of NO reacts with 0.5 mole of oxygen ionstead of 0.467 mole. Hence, it is clearly seen that there is less oxygen involved indicating that oxygen is the limiting reactant.

To calculate the number of moles of products

  $\begin{array}{c}{\text{2NO + O}}_2\to 2{\text{ NO}}_2\\ {\text{2 mole of NO=2 mole of NO}}_2\\ 0.933\text{ mole of NO=0}{\text{.933 mole of NO}}_2\end{array}$

Therefore, 0.933 mole of products is formed

To calculate the number of grams of the product

Molar mass of ${\text{NO}}_2$ = 46.01 g/mol

  $\begin{array}{c}{\text{1 mole of NO}}_2=46.01g\\ 0.93{\text{3 mole of NO}}_2=0.933\text{ x 46}\text{.01}\\ \text{=42}\text{.927g}\\ \approx \text{42}\text{.93g}\end{array}$