Chapter 9, Problem 60A

(a)

Interpretation Introduction

Interpretation:

The limiting reactant and the theoretical yield in grams of carbon dioxide need to be determined.

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(\text{l}\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(l\right)$

Concept Introduction:

The first step is to have a balanced chemical reaction. It is a necessity in every stoichiometric calculation.

After that, determine the number moles present in each of the reactant by simply dividing the given mass by the molar mass of the reactant. After obtaining the number of moles of the reactant, using the balance chemical reaction, relate the number of moles of the reactants to the moles of the product

Answer to Problem 60A

The balanced reaction is:

  $C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

limiting reagent is O₂(g)

Theoretical yield of CO₂=45.83 g

Explanation of Solution

The balanced reaction is as follows:

  $C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

1 mole of C₂H₅OH(l) reacts with 3 moles of O₂(g) to form 2 moles of CO₂.

  $\begin{array}{l}25.0gof{C}_2{H}_{5}OH=\frac{25.0}{46.07}=0.543moles\hfill \\ 25.0gof{O}_2=\frac{25.0}{16.0}=1.562moles\hfill \\ 0.543moles{C}_2{H}_{5}OHrequired0.543\times 3=1.629molesof{O}_2\hfill \end{array}$

Here, limiting reagent is O₂(g).

  $\begin{array}{l}molesofC{O}_{2}formed=2\times \frac{1.562}{3}=1.0413moles\hfill \\ massofC{O}_2=1.0413\times 44.01=45.83g\hfill \end{array}$

(b)

Interpretation Introduction

The limiting reactant and the theoretical yield in grams of nitrogen oxide need to be determined.

  $N_2\left(g\right)+O_2\left(g\right)\to NO\left(g\right)$

Interpretation:

Concept Introduction:

The first step is to have a balanced chemical reaction. It is a necessity in every stoichiometric calculation.

After that, determine the number moles present in each of the reactant by simply dividing the given mass by the molar mass of the reactant. After obtaining the number of moles of the reactant, using the balance chemical reaction, relate the number of moles of the reactants to the moles of the product

Answer to Problem 60A

The balanced reaction is

  $N_2\left(g\right)+O_2\left(g\right)\to 2NO\left(g\right)$

Limiting reagent is O2(g)

Theoretical yield of NO=93.75 g

Explanation of Solution

The balanced reaction is:

  $N_2\left(g\right)+O_2\left(g\right)\to 2NO\left(g\right)$

1 mole of N₂ reacts with 1 mole of O₂ to form 2 moles of NO.

  $\begin{array}{l}25.0gof{N}_2=25.0/14.0=1.786moles\hfill \\ 25.0gof{O}_2=25.0/16.0=1.562moles\hfill \end{array}$

Here, limiting reagent is O2(g)

  $\begin{array}{l}theoreticalyieldofNO=2\times 1.562=3.124moles\hfill \\ massofNOyield=3.124\times 30.01=93.75g\hfill \end{array}$

(c)

Interpretation Introduction

Interpretation:

The limiting reactant and the theoretical yield in grams of sodium chloride need 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:

The first step is to have a balanced chemical reaction. It is a necessity in every stoichiometric calculation.

After that, determine the number moles present in each of the reactant by simply dividing the given mass by the molar mass of the reactant. After obtaining the number of moles of the reactant, using the balance chemical reaction, relate the number of moles of the reactants to the moles of the product

Answer to Problem 60A

The balanced reaction is:

  $2NaCl{O}_2\left(aq\right)+C{l}_2\left(g\right)\to 2Cl{O}_2\left(g\right)+2NaCl\left(aq\right)$

limiting reagent is NaClO₂(aq)

Theoretical yield of NaCl=16.15 g

Explanation of Solution

The balanced reaction is:

  $2NaCl{O}_2\left(aq\right)+C{l}_2\left(g\right)\to 2Cl{O}_2\left(g\right)+2NaCl\left(aq\right)$

2 moles of NaClO2(aq) reacts with 1 mole of Cl₂(g) to form 2 moles of NaCl.

  $\begin{array}{l}25.0gNaCl{O}_2=\frac{25.0}{90.44}=0.2764mole\hfill \\ 25.0gC{l}_2\left(g\right)=\frac{25.0}{70.906}=0.3526moles\hfill \end{array}$

limiting reagent is NaClO₂(aq)

  $\begin{array}{l}theoreticalyieldofNaCl=0.2764moles\hfill \\ massofNaClyield=0.2764\times 58.44=16.15g\hfill \\ \hfill \end{array}$

(d)

Interpretation Introduction

Interpretation:

The limiting reactant and the theoretical yield in grams of ammonia need to be determined.

  $H_2\left(g\right)+N_2\left(g\right)\to N{H}_3\left(g\right)$

Concept Introduction:

The first step is to have a balanced chemical reaction. It is a necessity in every stoichiometric calculation.

After that, determine the number moles present in each of the reactant by simply dividing the given mass by the molar mass of the reactant. After obtaining the number of moles of the reactant, using the balance chemical reaction, relate the number of moles of the reactants to the moles of the product

Answer to Problem 60A

The balanced reaction is as follows:

  $3H_2\left(g\right)+N_2\left(g\right)\to 2N{H}_3\left(g\right)$

limiting reagent is N2(g)

theoretical yield of NH3=60.83 g

Explanation of Solution

The balanced reaction is as follows:

  $3H_2\left(g\right)+N_2\left(g\right)\to 2N{H}_3\left(g\right)$

3 moles of H₂ reacts with 1 mole of N₂ to form 2 moles of NH₃

  $\begin{array}{l}25.0gof{H}_2=25.0/2.02=12.376mole\hfill \\ 25.0gof{N}_2=\frac{25.0}{14.0}=1.786moles\hfill \\ here,limitingreagentis{N}_2\left(g\right)\hfill \\ theoreticalyieldofN{H}_3=2\times 1.786=3.572moles\hfill \\ massofN{H}_{3}formed=3.572\times 17.03=60.83g\hfill \end{array}$