Chapter 9, Problem 9A

(a)

Interpretation Introduction

Interpretation:

The number of moles and grams of ammonium chloride produced from 0.5 mole of ammonia must be calculated.

Concept Introduction:

One mole ammonia gas reacts with one mole hydrogen chloride gas to produce one mole of ammonium chloride with molar mass 53.5 g.

Answer to Problem 9A

0.5 mole of ammonium chloride will be produced. Mass of 0.5 mole ammonium chloride is 26.75 g.

Mole ratio of NH₃:NH₄Cl is 1:1.

Explanation of Solution

The given reaction is as follows:

  ${\text{NH}}_{\text{3}}\text{(g)+HCl(g)}\to {\text{NH}}_{\text{4}}\text{Cl(s)}\text{.}$

From the balanced equation it is clear that complete conversion of 0.5 mole ammonia will produce 0.5 mole ammonium chloride.

Molar mass of ammonium chloride is 53.5 g.

Thus mass of 0.5 mole ammonium chloride $\text{=0}\text{.5×53}\text{.5 g =26}\text{.75 g}\text{.}$

b)

Interpretation Introduction

Interpretation:

The number of moles and mass of carbon disulfide and hydrogen sulfide produced from 0.5 mole of atomic sulfur must be calculated.

Concept Introduction:

Four mole sulfur atom reacts with one mole methane gas to produce one mole of carbon disulfide with molar mass 76 g and 2 moles of hydrogen sulfide gas with molar mass 34 g.

Answer to Problem 9A

0.125 mole of carbon disulfide will be produced. Mass of 0.125 mole carbon disulfide is 9.5 g.

0.25 moles of hydrogen sulfide will be produced. Mass of 0.25 moles of hydrogen sulfide is 8.5 g.

Mole ratio of S: CS₂ is 4:1 and mole ratio of S: H₂S is 2:1.

Explanation of Solution

The given reaction is as follows:

  ${\text{CH}}_{\text{4}}\text{(g)+4S(s)}\to {\text{CS}}_{\text{2}}{\text{(l)+2H}}_{\text{2}}\text{S(g)}\text{.}$

From the balanced equation it is clear that complete conversion of 0.5 mole sulfur will produce 0.125 mole carbon disulfide (CS₂) as 4 moles S produces 1 mole CS₂.

Molar mass of carbon disulfide is 76 g.

Thus mass of 0.125 mole of carbon disulfide $\text{=0}\text{.125×76 g =9}\text{.5 g}\text{.}$

4 moles S produces 2 moles H₂S. Thus 0.5 mol S will produce 0.25 mole of H₂S.

Molar mass of hydrogen sulfide is 34 g.

Thus mass of 0.25 mole of hydrogen sulfide $\text{=0}\text{.25×34 g =8}\text{.5 g}\text{.}$

c)

Interpretation Introduction

Interpretation:

The number of moles and mass of phosphorous acid and hydrochloric acid produced from 0.5 mole of phosphorous trichloride must be calculated.

Concept Introduction:

One mole phosphorous acid reacts with three moles of water to produce one mole of phosphorous acid with molar mass 82 g and three moles of hydrochloric acid with molar mass 36.5 g.

Answer to Problem 9A

0.5 mole of phosphorous acid will be produced. Mass of 0.5 mole phosphorous acid is 41 g.

1.5 moles of hydrochloric acid will be produced. Mass of 1.5 moles of hydrochloric acid is 54.75 g.

Mole ratio of PCl₃:H₃PO₃ is 1:1 and mole ratio of PCl₃: HCl is 1:3.

Explanation of Solution

The reaction is as follows:

  ${\text{PCl}}_{\text{3}}{\text{(l)+3H}}_{\text{2}}\text{O(l)}\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{(aq)+3HCl(aq)}\text{.}$

From the balanced equation it is clear that complete conversion of 0.5 mole PCl₃ will produce 0.5 mole phosphorous acid (PCl₃).

Molar mass of H₃PO₃ is 82 g.

Thus mass of 0.5 mole H₃PO₃$\text{=0}\text{.5×82 g =41 g}\text{.}$

1 moles PCl₃ produces 3 moles HCl. Thus 0.5 mol PCl₃ will produce 1.5 mole of HCl.

Molar mass of HCl is 36.5 g.

Thus mass of 1.5 moles of HCl $\text{=1}\text{.5×36}\text{.5 g =54}\text{.75 g}\text{.}$

d)

Interpretation Introduction

Interpretation:

The number of moles and grams of sodium bicarbonate produced from 0.5 mole of sodium hydroxide must be calculated.

Concept Introduction:

One mole sodium hydroxide reacts with one mole of carbon dioxide to produce one mole of sodium bicarbonate with molar mass 84 g.

Answer to Problem 9A

0.5 mole of sodium bicarbonate will be produced. Mass of 0.5 mole sodium bicarbonate is 42 g.

Mole ratio of NaOH: NaHCO₃ is 1:1.

Explanation of Solution

The reaction is as follows:

  ${\text{NaOH(s)+CO}}_{\text{2}}\text{(g)}\to {\text{NaHCO}}_{\text{3}}\text{(s)}\text{.}$

From the balanced equation it is clear that complete conversion of 0.5 mole NaOH will produce 0.5 mole NaHCO₃

Molar mass of NaHCO₃ is 84 g.

Thus mass of 0.5 mole NaHCO₃$\text{=0}\text{.5×84 g =42 g}\text{.}$