Chapter 12, Problem 47A

a.

Interpretation Introduction

Interpretation: To calculate the amount of fluorine 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 47A

The amount of fluorine is $372\text{ g}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  $5{\text{F}}_2\left(g\right)+2{\text{NH}}_3\left(g\right)\to {\text{N}}_2{\text{F}}_4\left(g\right)+6\text{ HF}\left(g\right)$

When five moles of fluorine react with two moles of ammonia, it yields one mole of tetrafluoro hydrazine and six moles of hydrogen fluoride.

The amount of ammonia is $6.66\text{ g}$ .

The molar mass of ammonia is ${\text{17 g mol}}^{-1}$ .

The molar mass of fluorine is ${\text{38 g mol}}^{-1}$

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

  $6.66\cancel{{\text{g NH}}_3}\times \frac{1\cancel{{\text{mol NH}}_3}}{\text{17 }\cancel{{\text{g NH}}_3}}\times \frac{5\cancel{{\text{mol F}}_2}}{2\cancel{{\text{mol NH}}_3}}\times \frac{{\text{38 g F}}_2}{1\cancel{{\text{mol F}}_2}}=372{\text{ g F}}_2$

b.

Interpretation Introduction

Interpretation: To calculate the amount of ammonia 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. In a balanced chemical equation, the coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 47A

The amount of ammonia is $\text{1}\text{.32 g}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  $5{\text{F}}_2\left(g\right)+2{\text{NH}}_3\left(g\right)\to {\text{N}}_2{\text{F}}_4\left(g\right)+6\text{ HF}\left(g\right)$

When five moles of fluorine react with two moles of ammonia, it yields one mole of tetrafluoro hydrazine and six moles of hydrogen fluoride.

The amount of hydrogen fluoride is $\text{4}\text{.65 g}$ .

The molar mass of ammonia is $\text{20}{\text{.0 g mol}}^{-1}$ .

The molar mass of hydrogen fluoride is ${\text{38 g mol}}^{-1}$

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

  $4.65\cancel{\text{g HF}}\times \frac{1\cancel{\text{mol HF}}}{20.0\cancel{\text{g HF}}}\times \frac{2\cancel{{\text{mol NH}}_3}}{\text{6}\cancel{\text{mol HF}}}\times \frac{{\text{17 g NH}}_3}{1\cancel{{\text{mol NH}}_3}}=1.32{\text{ g NH}}_3$

c.

Interpretation Introduction

Interpretation: To calculate the grams of tetrafluoro hydrazine.

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. In a balanced chemical equation, the coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 47A

The amount of tetrafluoro hydrazine is $123\text{ g}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  $5{\text{F}}_2\left(g\right)+2{\text{NH}}_3\left(g\right)\to {\text{N}}_2{\text{F}}_4\left(g\right)+6\text{ HF}\left(g\right)$

When five moles of fluorine react with two moles of ammonia, it yields one mole of tetrafluoro hydrazine and six moles of hydrogen fluoride.

The amount of fluorine is $\text{225 g}$ .

The molar mass of tetrafluoro hydrazine is ${\text{104 g mol}}^{-1}$ .

The molar mass of fluorine is ${\text{38 g mol}}^{-1}$ .

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

  $225\cancel{{\text{g F}}_2}\times \frac{1\cancel{{\text{mol F}}_2}}{\text{38}\cancel{{\text{g F}}_2}}\times \frac{1\cancel{{\text{mol N}}_2{\text{F}}_4}}{5\cancel{{\text{mol F}}_2}}\times \frac{{\text{104 g N}}_2{\text{F}}_4}{1\cancel{{\text{mol N}}_2{\text{F}}_4}}=123{\text{ g N}}_2{\text{F}}_4$