Q4.112CP - Chemistry: Molecular Nature Of Matter And Change

Question

The brewing industry uses yeast microorganisms to convert glucose to ethanol for wine and beer. The baking industry uses the carbon dioxide produced to make bread rise:

 

 ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{6}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\stackrel{\mathbf{yeast}}{\mathbf{\to }}\mathbf{2}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

 

How many grams of ethanol can be produced from 100g of glucose? What volume of CO₂ is produced? (Assume 1 mol of gas occupies 22.4 L at the conditions used.)

Step-by-Step Solution

Answer

You need to calculate how many grams of ethanol can be produced from 100g of glucose. Also you need to calculate volume of CO₂ produced.

Step 1: Balanced chemical equation

The brewing industry uses yeast microorganisms to convert glucose to ethanol for wine and beer. The baking industry uses the carbon dioxide produced to make bread rise:

 

 ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)\stackrel{\mathrm{yeast}}{\to }2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+2{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Step 2: Calculation of moles of C 6 H 12 O 6

Mass of C₆H₁₂O₆ = 100g

 

Molecular mass of C₆H₁₂O₆ = 180.096g/mol

 

Again you know,

 $\mathrm{moles}=\frac{\mathrm{mass}\left(\mathrm{given}\right)}{\mathrm{mass}\left(\mathrm{molar}\right)}$

 

Moles of C₆H₁₂O₆

 _{$\begin{array}{l}=\frac{100}{180.096}\left(\mathrm{mol}\right)\\ =0.555\left(\mathrm{mol}\right)\end{array}$}

 

 

Hence, moles of C₆H₁₂O₆ are 0.555mol.

Step 3: Calculation of moles of C 2 H 5 OH

According to the balanced equation

 

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)\stackrel{\mathrm{yeast}}{\to }2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+2{\mathrm{CO}}_2\left(\mathrm{g}\right)$

 

1 mole of C₆H₁₂O₆ produces 2 moles C₂H₅OH

Now, moles of C₂H₅OH

 ^{$\begin{array}{l}=0.555\times \frac{2}{1}\left(\mathrm{mol}\right)\\ =1.11\left(\mathrm{mol}\right)\end{array}$}

 

 Hence, moles of C₂H₅OH are 1.11mol.

Step 4: Calculation of mass of C 2 H 5 OH

Moles of C₂H₅OH = 1.11moles

 

Molecular mass of C₂H₅OH = 46.048g/mol

 

Again you know,

 $\mathrm{mass}=\mathrm{moles}\times \mathrm{mass}\left(\mathrm{molar}\right)$

 

Mass of C₂H₅OH

 _{$\begin{array}{l}=1.11\times 46.048\left(\mathrm{g}\right)\\ =51.113\left(\mathrm{g}\right)\end{array}$}

_{Hence, mass of C2H5OH is 51.113g.} 

Step 5: Calculation of moles of CO 2

According to the balanced equation

 

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)\stackrel{\mathrm{yeast}}{\to }2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+2{\mathrm{CO}}_2\left(\mathrm{g}\right)$

 

1 mole of C₆H₁₂O₆ produces 2 moles CO₂

 

Now, moles of CO₂

 ^{$\begin{array}{l}=0.555\times \frac{2}{1}\left(\mathrm{mol}\right)\\ =1.11\left(\mathrm{mol}\right)\end{array}$}

 

 

Hence, moles of CO₂ are 1.11mol.

Step 6: Calculation of volume of CO 2

Again you can write,

 

Volume of CO₂

 $\begin{array}{l}=\left[\mathrm{moles}\left({\mathrm{CO}}_2\right)\times \frac{22.4\left(\mathrm{L}\right)}{1\left(\mathrm{mole}\right)}\right]\left(\mathrm{L}\right)\\ =\left(1.11\times \frac{22.4}{1}\right)\left(\mathrm{L}\right)\\ =24.864\left(\mathrm{L}\right)\end{array}$

 

 

Hence, volume of CO₂ is 24.864L.

 

Step 7: Conclusion

Hence, 51.113 grams of ethanol can be produced from 100g of glucose and the volume of CO₂ produced is 24.864L.