91Ó°ÊÓ

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:

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

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.

According to the balanced equation

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

1 mole of C₆H₁₂O₆ produces 2 moles C₂H₅OH

Now, moles of C₂H₅OH

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

Hence, moles of C₂H₅OH are 1.11mol.

Moles of C₂H₅OH = 1.11moles

Molecular mass of C₂H₅OH = 46.048g/mol

Again you know,

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

Mass of C₂H₅OH

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

_{Hence, mass of C2H5OH is 51.113g.}

According to the balanced equation

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

1 mole of C₆H₁₂O₆ produces 2 moles CO₂

Now, moles of CO₂

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

Hence, moles of CO₂ are 1.11mol.

Again you can write,

Volume of CO₂

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

Hence, volume of CO₂ is 24.864L.

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