# What mass of calcium carbonate could be collected, if a 90*g mass of glucose is fermented, and the product gases are vented to a solution of EXCESS limewater?

Aug 23, 2016

We need a stoichiometric equation for the fermentation of glucose. Approx. $100 \cdot g$ calcium carbonate could be collected.

#### Explanation:

${C}_{6} {H}_{12} {O}_{6} \rightarrow 2 {C}_{2} {H}_{5} O H + 2 C {O}_{2}$

One mole of glucose ferments to give 2 mole of ethyl alcohol, and 2 moles of carbon dioxide gas. When you do your brewing at home, you can often judge the end of the fermentation by the fact that the bubbler (i.e. the air lock on your fermenter) stops bubbling carbon dioxide gas.

$\text{Moles of glucose}$ $=$ $\frac{90 \cdot g}{180.16 \cdot g \cdot m o {l}^{-} 1}$ $\cong$ $\frac{1}{2} \cdot m o l$.

By the stoichiometry of the fermentation reaction, $1$ $m o l$ of carbon dioxide would be evolved. We need another equation to show the reaction with limewater, $C a {\left(O H\right)}_{2}$:

$C a {\left(O H\right)}_{2} \left(a q\right) + C {O}_{2} \left(g\right) \rightarrow C a C {O}_{3} \left(s\right) \downarrow + {H}_{2} O \left(l\right)$

So at most, $1$ $m o l$ calcium carbonate could be collected:

$\text{Mass of calcium carbonate}$ $=$ $1 \cdot m o l \times 100.09 \cdot g \cdot m o {l}^{-} 1$.

What mass of ethanol is associated with the formation of this quantity?