Chapter 12, Problem 50A

a.

Interpretation Introduction

Interpretation: To calculate the amount of water in grams.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 50A

The amount of water is $51.2\text{ g}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  ${\text{Li}}_3\text{N}\left(s\right)+3{\text{H}}_2\text{O}\left(l\right)\to {\text{NH}}_3\left(g\right)+3\text{LiOH}\left(aq\right)$

When one mole of lithium nitride reacts with three moles of water, it yields one mole of ammonia and three moles of lithium hydroxide.

The amount of lithium nitride is $32.9\text{ g}$ .

The molar mass of lithium nitride is $\text{34}{\text{.7 g mol}}^{-1}$ .

The molar mass of water is ${\text{18 g mol}}^{-1}$ .

Therefore, the grams of water can be calculated as follows:

  $32.9\cancel{{\text{g Li}}_3\text{N}}\times \frac{1\cancel{{\text{mol Li}}_3\text{N}}}{34.7\cancel{{\text{g Li}}_3\text{N}}}\times \frac{3\cancel{{\text{mol H}}_2\text{O}}}{1\cancel{{\text{mol Li}}_3\text{N}}}\times \frac{{\text{18 g H}}_2\text{O}}{1\cancel{{\text{mol H}}_2\text{O}}}=51.2{\text{ g H}}_2\text{O}$

b.

Interpretation Introduction

Interpretation: To calculate the number of molecules of ammonia.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 50A

The number of molecules of ammonia are $1.132\times {10}^{23}\text{molecules}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  ${\text{Li}}_3\text{N}\left(s\right)+3{\text{H}}_2\text{O}\left(l\right)\to {\text{NH}}_3\left(g\right)+3\text{LiOH}\left(aq\right)$

When one mole of lithium nitride reacts with three moles of water, it yields one mole of ammonia and three moles of lithium hydroxide.

The amount of lithium nitride is $32.9\text{ g}$ .

The molar mass of lithium nitride is $\text{34}{\text{.7 g mol}}^{-1}$ .

Therefore, the molecules of ammonia can be calculated as follows:

  $32.9\cancel{{\text{g Li}}_3\text{N}}\times \frac{1\cancel{{\text{mol Li}}_3\text{N}}}{34.7\cancel{{\text{g Li}}_3\text{N}}}\times \frac{1\cancel{{\text{mol NH}}_3}}{1\cancel{{\text{mol Li}}_3\text{N}}}\times \frac{6.022\times {10}^{23}{\text{ NH}}_3\text{molecules}}{1\cancel{{\text{mol NH}}_3}}=5.71\times {10}^{23}{\text{ NH}}_3\text{molecules}$

c.

Interpretation Introduction

Interpretation: To calculate the amount of lithium nitride.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 50A

The grams of lithium nitride are $23.2\text{ g }$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  ${\text{Li}}_3\text{N}\left(s\right)+3{\text{H}}_2\text{O}\left(l\right)\to {\text{NH}}_3\left(g\right)+3\text{LiOH}\left(aq\right)$

When one mole of lithium nitride reacts with three moles of water, it yields one mole of ammonia and three moles of lithium hydroxide.

The liters of ammonia is $15.0\text{ L}$ .

Therefore, the molecules of ammonia can be calculated as follows:

  $15.0\cancel{{\text{L NH}}_3}\times \frac{1\cancel{{\text{mol NH}}_3}}{\text{22}\text{.4 }\cancel{{\text{L NH}}_3}}\times \frac{1\cancel{{\text{mol Li}}_{\text{3}}\text{N}}}{1\cancel{{\text{mol NH}}_3}}\times \frac{34.7{\text{ g Li}}_{\text{3}}\text{N }}{1\cancel{{\text{mol Li}}_{\text{3}}\text{N}}}=23.2{\text{ g Li}}_{\text{3}}\text{N}$