Chapter 6, Problem 6.63EP

Using each of the following balanced chemical equations, calculate the number of moles of the first listed reactant needed to produce 5.00 moles of CO₂.

1. a. ${\text{C}}_7{\text{H}}_{16}+11{\text{O}}_2\to 7{\text{CO}}_2+8{\text{H}}_2\text{O}$
2. b. ${\text{CS}}_2+{\text{O}}_2\to 2{\text{SO}}_2+{\text{CO}}_2$
3. c. ${\text{Fe}}_3{\text{O}}_4+\text{CO}\to \text{3FeO}\text{+}{\text{CO}}_2$
4. d. $2\text{HCl}\text{+}{\text{CaCO}}_3\to {\text{CaCl}}_2+{\text{CO}}_2+{\text{H}}_2\text{O}$

Interpretation Introduction

Interpretation:

The number of moles of $C_7{H}_{16}$ that is required to produce $5.00moles$ of $C{O}_2$ has to be calculated if the balanced chemical equation is $C_7{H}_{16}+11O_2\stackrel{}{\to }7C{O}_2+8H_{2}O$.

Concept Introduction:

The coefficients that are present in a balanced chemical equation is like subscripts in the chemical formula.  These coefficients can be interpreted in two levels.  One is a microscopic level and another is macroscopic level.

Microscopic level:

The coefficients that are present in a balanced chemical equation gives numerical relationship between the formula units consumed or produced when a chemical reaction takes place.

Macroscopic level:

In a balanced chemical equation, the coefficients gives information about the fixed molar ratios among the substance that is consumed or produced when a chemical reaction takes place.

Explanation of Solution

Given chemical equation is,

$C_7{H}_{16}+11O_2\stackrel{}{\to }7C{O}_2+8H_{2}O$

From the above chemical equation, it is found that $1.0mole$ of $C_7{H}_{16}$ reacts with $11.0moles$ of $O_2$ to form $7.0moles$ of $C{O}_2$ and $8.0moles$ of $H_{2}O$.  The moles of $C_7{H}_{16}$ that will be required to produce $5.00moles$ of $C{O}_2$ can be calculated using the conversion factor,

$\frac{\text{1}\text{.0}\text{mole}{\text{C}}_{\text{7}}{\text{H}}_{\text{16}}}{\text{7}\text{.0}\text{moles}{\text{CO}}_{\text{2}}}$

Number of moles of $C_7{H}_{16}$ required is,

  $5.00molesC{O}_{2}x\left(\frac{\text{1}\text{.0}\text{mole}{\text{C}}_{\text{7}}{\text{H}}_{\text{16}}}{\text{7}\text{.0}\text{moles}{\text{CO}}_{\text{2}}}\right)=0.714mole{C}_7{H}_{16}$

Therefore, $0.714mole$ of $C_7{H}_{16}$ will be required to produce $5.00moles$ of $C{O}_2$.

Conclusion

The number of moles of $C_7{H}_{16}$ that is required to produce $5.00moles$ of $C{O}_2$ is calculated.

Interpretation Introduction

Interpretation:

The number of moles of $C{S}_2$ that is required to produce $5.00moles$ of $C{O}_2$ has to be calculated if the balanced chemical equation is $C{S}_2+O_2\stackrel{}{\to }2S{O}_2+C{O}_2$.

Concept Introduction:

The coefficients that are present in a balanced chemical equation is like subscripts in the chemical formula.  These coefficients can be interpreted in two levels.  One is a microscopic level and another is macroscopic level.

Microscopic level:

The coefficients that are present in a balanced chemical equation gives numerical relationship between the formula units consumed or produced when a chemical reaction takes place.

Macroscopic level:

In a balanced chemical equation, the coefficients gives information about the fixed molar ratios among the substance that is consumed or produced when a chemical reaction takes place.

Explanation of Solution

Given chemical equation is,

$C{S}_2+O_2\stackrel{}{\to }2S{O}_2+C{O}_2$

From the above chemical equation, it is found that $1.0mole$ of $C{S}_2$ reacts with $1.0mole$ of $O_2$ to form $1.0mole$ of $C{O}_2$ and $2.0moles$ of $S{O}_2$.  The moles of $C{S}_2$ that will be required to produce $5.00moles$ of $C{O}_2$ can be calculated using the conversion factor,

$\frac{\text{1}\text{.0}\text{mole}C{S}_2}{\text{1}\text{.0}\text{mole}{\text{CO}}_{\text{2}}}$

Number of moles of $C{S}_2$ required is,

  $5.00molesC{O}_{2}x\left(\frac{\text{1}\text{.0}\text{mole}C{S}_2}{\text{1}\text{.0}\text{mole}{\text{CO}}_{\text{2}}}\right)=5.00molesC{S}_2$

Therefore, $5.00moles$ of $C{S}_2$ will be required to produce $5.00moles$ of $C{O}_2$.

Conclusion

The number of moles of $C{S}_2$ that is required to produce $5.00moles$ of $C{O}_2$ is calculated.

Interpretation Introduction

Interpretation:

The number of moles of $F{e}_3{O}_4$ that is required to produce $5.00moles$ of $C{O}_2$ has to be calculated if the balanced chemical equation is $F{e}_3{O}_4+CO\stackrel{}{\to }3FeO+C{O}_2$.

Concept Introduction:

The coefficients that are present in a balanced chemical equation is like subscripts in the chemical formula.  These coefficients can be interpreted in two levels.  One is a microscopic level and another is macroscopic level.

Microscopic level:

The coefficients that are present in a balanced chemical equation gives numerical relationship between the formula units consumed or produced when a chemical reaction takes place.

Macroscopic level:

In a balanced chemical equation, the coefficients gives information about the fixed molar ratios among the substance that is consumed or produced when a chemical reaction takes place.

Explanation of Solution

Given chemical equation is,

$F{e}_3{O}_4+CO\stackrel{}{\to }3FeO+C{O}_2$

From the above chemical equation, it is found that $1.0mole$ of $F{e}_3{O}_4$ reacts with $1.0mole$ of $CO$ to form $1.0mole$ of $C{O}_2$ and $3.0moles$ of $FeO$.  The moles of $F{e}_3{O}_4$ that will be required to produce $5.00moles$ of $C{O}_2$ can be calculated using the conversion factor,

$\frac{\text{1}\text{.0}\text{mole}F{e}_3{O}_4}{\text{1}\text{.0}\text{mole}{\text{CO}}_{\text{2}}}$

Number of moles of $F{e}_3{O}_4$ required is,

  $5.00molesC{O}_{2}x\left(\frac{\text{1}\text{.0}\text{mole}F{e}_3{O}_4}{\text{1}\text{.0}\text{mole}{\text{CO}}_{\text{2}}}\right)=5.00molesF{e}_3{O}_4$

Therefore, $5.00moles$ of $F{e}_3{O}_4$ will be required to produce $5.00moles$ of $C{O}_2$.

Conclusion

The number of moles of $F{e}_3{O}_4$ that is required to produce $5.00moles$ of $C{O}_2$ is calculated.

Interpretation Introduction

Interpretation:

The number of moles of $HCl$ that is required to produce $5.00moles$ of $C{O}_2$ has to be calculated if the balanced chemical equation is $2HCl+CaC{O}_3\stackrel{}{\to }CaC{l}_2+C{O}_2+H_{2}O$.

Concept Introduction:

The coefficients that are present in a balanced chemical equation is like subscripts in the chemical formula.  These coefficients can be interpreted in two levels.  One is a microscopic level and another is macroscopic level.

Microscopic level:

The coefficients that are present in a balanced chemical equation gives numerical relationship between the formula units consumed or produced when a chemical reaction takes place.

Macroscopic level:

In a balanced chemical equation, the coefficients gives information about the fixed molar ratios among the substance that is consumed or produced when a chemical reaction takes place.

Explanation of Solution

Given chemical equation is,

$2HCl+CaC{O}_3\stackrel{}{\to }CaC{l}_2+C{O}_2+H_{2}O$

From the above chemical equation, it is found that $2.0moles$ of $HCl$ reacts with $1.0mole$ of $CaC{O}_3$ to form $1.0mole$ of $C{O}_2$, $1.0mole$ of $CaC{l}_2$ and $1.0mole$ of $H_{2}O$.  The moles of $HCl$ that will be required to produce $5.00moles$ of $C{O}_2$ can be calculated using the conversion factor,

$\frac{\text{2}\text{.0}\text{moles}HCl}{\text{1}\text{.0}\text{mole}{\text{CO}}_{\text{2}}}$

Number of moles of $HCl$ required is,

  $5.00molesC{O}_{2}x\left(\frac{\text{2}\text{.0}\text{moles}HCl}{\text{1}\text{.0}\text{mole}{\text{CO}}_{\text{2}}}\right)=10.00molesHCl$

Therefore, $10.00moles$ of $HCl$ will be required to produce $5.00moles$ of $C{O}_2$.

Conclusion

The number of moles of $HCl$ that is required to produce $5.00moles$ of $C{O}_2$ is calculated.