Some organisms use fermentation to break down glucose for energy. The following equation represents the process of fermentation, where glucose is converted into ethanol and carbon dioxide.

$$ \ce { C6H12O6(s) -> 2 C2H5OH(l) + 2 CO2(g) } $$

Calculate the mass of ethanol produced if 100.0 grams of glucose reacts completely.

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Start by writing down the given value and units.

$$ \pu{100.0 g } \ce{ C6H12O6} $$ $$ \pu{100.0 g } \ce{ C6H12O6} \cdot \colorbox{yellow}{ \( \frac{\pu{1 mol } { \ce{C6H12O6}}}{\pu{180.2 g } \ce{ C6H12O6}}\) } \tag*{\tiny convert mass into moles}$$ $$ \pu{100.0 g } \ce{ C6H12O6} \cdot \frac{\pu{1 mol } { \ce{C6H5OH}}}{\pu{180.2 g } \ce{ C6H12O6}} \cdot \colorbox{yellow}{ \( \frac{\pu{2 mol } { \ce{C6H5OH}}}{\pu{1 mol } { \ce{C6H12O6}}} \)} \tag*{\tiny use mole ratio}$$ $$ \pu{100.0 g } \ce{ C6H12O6} \cdot \frac{\pu{1 mol } { \ce{C6H5OH}}}{\pu{180.2 g } \ce{ C6H12O6}} \cdot \frac{\pu{2 mol } { \ce{C6H5OH}}}{\pu{1 mol } { \ce{C6H12O6}}} \cdot \colorbox{yellow} { \( \frac{\pu{46.1 g } \ce{ C6H5OH}}{\pu{1 mol } { \ce{C6H5OH}}} \) } \tag*{\tiny convert to mass}$$ $$ \pu{100.0 }\cancel{\pu{ g } \ce{ C6H12O6}} \cdot \frac{ \pu{ 1 } \cancel{ \pu{ mol } \ce{C6H5OH}}}{\pu{ 180.2 } \cancel{\pu{ g } \ce{ C6H12O6}}} \cdot \frac{ \pu{ 2 } \cancel {\pu { mol } \ce{C6H5OH}}}{\pu{ 1 } \cancel{\pu { mol } \ce{C6H12O6}}} \cdot \frac{\pu{46.1 g } \ce{ C6H5OH}}{\pu{ 1 } \cancel{\pu{ mol } \ce{C6H5OH}}} \tag*{\tiny check units}$$ $$ (100.0) \cdot \left(\frac{1}{180.2}\right) \cdot \left(\frac{2}{1}\right) \cdot \left(\frac{46.1}{1}\right) \pu { g } \ce {C6H5OH} \tag*{\tiny multiply through} $$ $$ \approx \pu{51.165 g } \ce {C6H5OH} \tag*{\tiny sig. figures}$$ $$ = \boxed{\pu {51.17 g } \ce {C6H5OH}}$$