Chapter 9, Problem 56A

(a)

Interpretation Introduction

Interpretation: The limiting reactant for the given reaction needs to be determined and theoretical yield of carbon dioxide needs to be determined.

  $C_2{H}_{5}O{H}_{ (l)} +O_{2(g)}\to C{O}_{2 }{}_{(g)} +H_{2}O{ }_{(l)}$

Concept Introduction:

A limiting reactant limits the amount of product depends on the reaction.

Answer to Problem 56A

.Limiting reagent => O₂

mass CO₂ = 22.92 g

Explanation of Solution

Balanced equation is as:

  $C_2{H}_{5}O{H}_{ \left(l\right)} +3O_{2\left(g\right)} \to 2 C{O}_{2 }{}_{\left(g\right)} +3H_{2}O{ }_{\left(l\right)}$

The number of moles of each reactant can be calculated as follows:

  $\begin{array}{l}moles=\frac{mass}{molarmass}\hfill \\ moles{C}_2{H}_{5}OH=\frac{25g}{46.07g/mole}=0.5427moles\hfill \\ moles{O}_2 =\frac{25g}{32g/mole}=0.7812moles\hfill \end{array}$

Here,

Mole Ratio C₂H₅OH : O₂ = 1:3

From moles of C₂H₅OH , moles O₂ needed for complete reaction should be

  $0.5427moles\times 3=1.648moles{O}_2$

However, available number of moles are 0.7812 moles O₂

Thus,

Limiting reagent has fewer moles hence stops the reaction.

Limiting reagent = O₂

Theoretical yield

Now, use limiting reagent

Mole Ratio O₂ : CO₂ = 3:2

Thus,

  $\begin{array}{l} molesC{O}_{2 }=\frac{2}{3}\times 0.7812=0.52083moles\hfill \\ mass=moles\times molarmass\hfill \\ massC{O}_{2 }=0.52083moles\times 44.01g/mole\hfill \\ massC{O}_{2 }=22.92g\hfill \end{array}$

(b)

Interpretation Introduction

Interpretation: The limiting reactant for the given reaction needs to be determined and theoretical yield of NO needs to be determined.

  $N_{2\left(g\right)} +O_{2\left(g\right)} \to N{O}_{ }{}_{\left(g\right)}$

Concept Introduction:

A limiting reactant limits the amount of product depends on the reaction.

Answer to Problem 56A

• Limiting reagent => O₂
• mass NO= 46.89 g

Explanation of Solution

Balanced equation is as:

  $N_{2\left(g\right)} +O_{2\left(g\right)} \to 2 N{O}_{ }{}_{\left(g\right)}$

The number of moles of N₂ and O₂ can be calculated as follows:

  $\begin{array}{l}moles{N}_2 =\frac{25g}{28g/mole}=0.89285moles\hfill \\ moles{O}_2 =\frac{25g}{32g/mole}=0.7812moles\hfill \end{array}$

Now,

Mole Ratio N₂ : O₂ = 1:1

Thus, moles of N₂ should be equal moles O₂

However, 0.7812 moles O₂ and 0.8928 moles N₂are present.

Thus limiting reagent has fewer moles hence stops the reaction.

Limiting reagent = O₂

Theoretical yield

Use limiting reagent

Mole Ratio O₂ : NO = 1:2

Thus,

  $\begin{array}{l}molesNO=2\times 0.7812moles=1.5625molesNO\hfill \\ mass=moles\times molarmass\hfill \\ massNO=1.5625moles\times 30.01g/mole\hfill \\ massNO=46.89g\hfill \end{array}$

(c)

Interpretation Introduction

Interpretation: The limiting reactant for the given reaction needs to be determined and theoretical yield of NaCl needs to be determined.

  $NaCl{O}_{2\left(aq\right)} +C{l}_{2\left(g\right)} \to Cl{O}_{2 \left(g\right)} + NaCl{ }_{\left(aq\right)}$

Concept Introduction:

A limiting reactant limits the amount of product depends on the reaction.

Answer to Problem 56A

• Limiting reagent => NaClO₂
• mass NaCl= 16.15 g

Explanation of Solution

Balanced equation is as:

  $\begin{array}{l}2NaCl{O}_{2\left(aq\right)} +C{l}_{2\left(g\right)} \to 2Cl{O}_{2 \left(g\right)} +2 NaCl{ }_{\left(aq\right)}\hfill \\ moles=\frac{mass}{molarmass}\hfill \\ molesNaCl{O}_2 =\frac{25g}{90.44g/mole}=0.2764moles\hfill \\ molesC{l}_2 =\frac{25g}{70.906g/mole}=0.35258moles\hfill \end{array}$

Mole Ratio NaClO₂ : Cl₂ = 2:1

Thus

From moles of Cl₂ , moles NaClO₂ needed for complete reaction should be twice the number of moles of Cl₂ that is 0.70516 mol.

However, available number of moles is 0.2764 moles NaClO₂

Thus

Limiting reagent has fewer moles hence stops the reaction.

Limiting reagent NaClO₂

Theoretical yield

Use limiting reagent

Mole Ratio NaClO₂ : NaCl = 2:2 = 1:1

Thus,

  $\begin{array}{l}{n}_{NaCl}=1\times 0.276=0.2764\text{ mol}\\ m=n\times M\\ =0.2764\times 58.44\\ =16.15\text{ g}\end{array}$

(d)

Interpretation Introduction

Interpretation: The limiting reactant for the given reaction needs to be determined and theoretical yield of ammonia needs to be determined.

  $H_{2\left(aq\right)} +N_{2\left(g\right)} \to N{H}_{3 }{}_{\left(g\right)}$

Concept Introduction:

A limiting reactant limits the amount of product depends on the reaction.

Answer to Problem 56A

• Limiting reagent => N₂
• mass NH₃= 30.41 g

Explanation of Solution

Balanced equation is as:

  $\begin{array}{l}3H_{2\left(aq\right)} +N_{2\left(g\right)} \to 2 N{H}_{3 }{}_{\left(g\right)}\hfill \\ moles=\frac{mass}{molarmass}\hfill \\ moles{H}_2 =\frac{25g}{2.01588g/mole}=12.40moles\hfill \\ moles{N}_2 =\frac{25g}{28g/mole}=0.89286moles\hfill \end{array}$

Mole Ratio H₂ : N₂ = 3:1

Thus

From moles of H₂ , moles N₂ needed for complete rxn should be:

  $n_{N_2}=\frac{12.40}{3}=4.134\text{ mol}$

However, available number of moles is 0.89286 moles N₂

Thus,

Limiting reagent has fewer moles hence stops the reaction.

Limiting reagent =N₂

Theoretical yield

Use limiting reagent

Mole Ratio N₂ : NH₃ = 1:2

  $\begin{array}{l}Thus,molesN{H}_3 =2\times 0.89286moles=1.7857molesN{H}_3\hfill \\ mass=molesxmolarmass\hfill \\ massN{H}_3 =1.7857moles\times 17.031g/mole\hfill \end{array}$

mass NH₃= 30.41 g