Chapter 9, Problem 45QAP

For each of the following unbalanced reactions, suppose exactly 5.00 g of each reactant is taken. Determine which reactant is limiting, and also determine what mass of the excess reagent will remain after the limiting reactant is consumed.

1. Na₂B₄O₇(s) + H₂SO₄(o
H ₃BOj(j) + Na ₂SO ₄(u
2. CaC,(s) + H₂O(/) → Ca(OH)₂(s) + C₂H₂(g)
3. NaCl(s) + H₂SO₄(/> → HCl(g) + Na₂SO₄(s)
4. SiO₂(s) + C(x) —> Si(/) + CO(g)

Interpretation Introduction

Interpretation : For the given reaction with 5 g of each reactant, the limiting reactant and the mass of the excess reactant after the completion of limiting reactant needs to be determined.

  ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{BO}}_{\text{3}}{\text{+Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

Concept Introduction : Stoichiometry is used in order to relate molar and mass quantities. The limiting reactant will always finish first than the excess materials.

Answer to Problem 45QAP

  $Limitingreac\tan t:N{a}_2{B}_4{O}_7$

  $Totalexcessmass=7.5212g$

Explanation of Solution

Given

  $m=5.00gofeachmaterial$

In order to make stoichiometric calculations, the reaction must be balanced:

  $N{a}_2{B}_4{O}_7+H_{2}S{O}_4+H_{2}O\to H_{3}B{O}_3+N{a}_{2}S{O}_4$

Balance B:

  $N{a}_2{B}_4{O}_7+H_{2}S{O}_4+H_{2}O\to 4H_{3}B{O}_3+N{a}_{2}S{O}_4$

Balance H

  $N{a}_2{B}_4{O}_7+H_{2}S{O}_4+5H_{2}O\to 4H_{3}B{O}_3+N{a}_{2}S{O}_4$

This is now balanced.

Calculate moles of reactants

  $\begin{array}{l}molN{a}_2{B}_4{O}_7=\frac{mass}{MM}=\frac{5.00g}{381.37g/mol}=0.0131\\ mol{H}_{2}S{O}_4=\frac{mass}{MM}=\frac{5.00g}{98.079g/mol}=0.051\\ mol{H}_{2}O=\frac{mass}{MM}=\frac{5.00g}{18.0g/mol}=0.277\end{array}$

Where MM is molar mass

Perform stoichiometric calculations:

  $\begin{array}{l}molN{a}_{2}S{O}_4=(0.0131molN{a}_2{B}_4{O}_7)\left(\frac{1molN{a}_{2}S{O}_4}{1molN{a}_2{B}_4{O}_7}\right)=0.0131\\ molN{a}_{2}S{O}_4=(0.051mol{H}_{2}S{O}_4)\left(\frac{1molN{a}_{2}S{O}_4}{1mol{H}_{2}S{O}_4}\right)=0.051\\ molN{a}_{2}S{O}_4=(0.277mol{H}_{2}O)\left(\frac{1molN{a}_{2}S{O}_4}{5mol{H}_{2}O}\right)=0.0555\end{array}$

From these calculations, the limiting reactant is $N{a}_2{B}_4{O}_7$ .

Excess moles:

  $\begin{array}{l}mol{H}_{2}S{O}_{4}left=0.051-0.0131=0.0379\\ mol{H}_{2}Oleft=(0.277-0.0131\times 5=0.277-0.0655=0.2115\end{array}$

Mass left:

  $\begin{array}{l}mass{H}_{2}S{O}_4=0.0379mol\left(98g/mol\right)=3.7142g\\ mass{H}_{2}Oleft=0.2115mol(18g/mol)=3.807g\\ Total=3.807+3.7142g\\ Totalexcessmass=7.5212g\end{array}$

Interpretation Introduction

Interpretation : For the given reaction with 5 g of each reactant, the limiting reactant and the mass of the excess reactant after the completion of limiting reactant needs to be determined.

  ${\text{CaC}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\to {\text{Ca(OH)}}_{\text{2}}{\text{+C}}_{\text{2}}{\text{H}}_{\text{2}}$

Concept Introduction : Stoichiometry is used in order to relate molar and mass quantities. The limiting reactant will always finish first than the excess materials.

Answer to Problem 45QAP

  $Limitingreac\tan t:Ca{C}_2$

  $Totalexcessmass=2.18628g$

Explanation of Solution

Given

  $m=5.00gofeachmaterial$

In order to make stoichiometric calculations, the reaction must be balanced:

  $Ca{C}_2+H_{2}O\to Ca{(OH)}_2+C_2{H}_2$

Balance O:

  $Ca{C}_2+2H_{2}O\to Ca{(OH)}_2+C_2{H}_2$

This is now balanced.

Calculate moles of reactants

  $\begin{array}{l}molCa{C}_2=\frac{mass}{MM}=\frac{5.00g}{64g/mol}=0.07812\\ mol{H}_{2}O=\frac{mass}{MM}=\frac{5.00g}{18.0g/mol}=0.277\end{array}$

Perform stoichiometric calculations:

  $\begin{array}{l}mol{C}_2{H}_2=(0.07812molCa{C}_2)\left(\frac{1mol{C}_2{H}_2}{1molCa{C}_2}\right)=0.07812\\ mol{C}_2{H}_2=(0.277mol{H}_{2}O)\left(\frac{1mol{C}_2{H}_2}{2mol{H}_{2}O}\right)=0.1385\end{array}$

From these calculations, the limiting reactant is $Ca{C}_2$ .

Excess moles:

  $mol{H}_{2}Oleft=(0.277)-(0.07812)(2)=0.12146mol$

Mass left:

  $\begin{array}{l}mass{H}_{2}Oleft=0.12146mol(18g/mol)=2.18628g\\ Totalexcessmass=2.18628g\end{array}$

Interpretation Introduction

Interpretation : For the given reaction with 5 g of each reactant, the limiting reactant and the mass of the excess reactant after the completion of limiting reactant needs to be determined.

  ${\text{NaCl+H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{HCl+Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

Concept Introduction : Stoichiometry is used in order to relate molar and mass quantities. The limiting reactant will always finish first than the excess materials.

Answer to Problem 45QAP

  $Limitingreac\tan t:NaCl$

  $Totalexcessmass=0.7742g$

Explanation of Solution

Given

  $m=5.00gofeachmaterial$

In order to make stoichiometric calculations, the reaction must be balanced:

  $NaCl+H_{2}S{O}_4\to HCl+N{a}_{2}S{O}_4$

Balance Na:

  $2NaCl+H_{2}S{O}_4\to HCl+N{a}_{2}S{O}_4$

Balance H:

  $2NaCl+H_{2}S{O}_4\to 2HCl+N{a}_{2}S{O}_4$

This is now balanced.

Calculate moles of reactants

  $\begin{array}{l}molNaCl=\frac{mass}{MM}=\frac{5.00g}{58g/mol}=0.0862\\ mol{H}_{2}S{O}_4=\frac{mass}{MM}=\frac{5.00g}{98g/mol}=0.0510\end{array}$

Perform stoichiometric calculations:

  $\begin{array}{l}molHCl=(0.0862molNaCl)\left(\frac{2molHCl}{2molNaCl}\right)=0.0862\\ molHCl=(0.0510mol{H}_{2}S{O}_4)\left(\frac{2molHCl}{1mol{H}_{2}S{O}_4}\right)=0.102\end{array}$

From these calculations, the limiting reactant is $NaCl$ .

Excess moles:

  $mol{H}_{2}S{O}_4=(0.051-0.0862x\frac{1}{2})=0.0079$

Mass left:

  $\begin{array}{l}mass{H}_{2}S{O}_{4}left=0.0079mol(98g/mol)=0.7742g\\ Totalexcessmass=0.7742g\end{array}$

Interpretation Introduction

Interpretation : For the given reaction with 5 g of each reactant, the limiting reactant and the mass of the excess reactant after the completion of limiting reactant needs to be determined.

Concept Introduction : Stoichiometry is used in order to relate molar and mass quantities. The limiting reactant will always finish first than the excess materials.

Answer to Problem 45QAP

  $Limitingreac\tan t:C$

  $Totalexcessmass=0.25g$

Explanation of Solution

Given

  $m=5.00gofeachmaterial$

In order to make stoichiometric calculations, the reaction must be balanced:

  $Si{O}_2+C\to Si+CO$

Balance O:

  $Si{O}_2+C\to Si+2CO$

Balance C:

  $Si{O}_2+2C\to Si+2CO$

This is now balanced.

Calculate moles of reactants

  $\begin{array}{l}molSiO2=\frac{mass}{MM}=\frac{5.00g}{60.08g/mol}=0.0833\\ molC=\frac{mass}{MM}=\frac{5.00g}{12g/mol}=0.4166\end{array}$

Perform stoichiometric calculations:

  $\begin{array}{l}molSi=(0.0833molNaCl)\left(\frac{1molSi}{1molSi{O}_2}\right)=0.0862\\ molSi=(0.4166mol{H}_{2}S{O}_4)\left(\frac{1molSi}{2molC}\right)=0.2083\end{array}$

From these calculations, the limiting reactant is $Si{O}_2$ .

Excess moles:

  $molC=0.4166-(0.0833)x2=0.25$

Mass left:

  $\begin{array}{l}massCleft=0.25mol(12g/mol)=3.0g\\ Totalexcessmass=3.0g\end{array}$