Question

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?

Answer 1

We need a stoichiometric equation for the fermentation of glucose. Approx. `100*g` calcium carbonate could be collected.

Explanation:

`C_6H_12O_6 rarr 2C_2H_5OH + 2CO_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.

`"Moles of glucose"` `=` `(90*g)/(180.16*g*mol^-1)` `~=` `1/2*mol`.

By the stoichiometry of the fermentation reaction, `1` `mol` of carbon dioxide would be evolved. We need another equation to show the reaction with limewater, `Ca(OH)_2`:

`Ca(OH)_2(aq) + CO_2(g) rarr CaCO_3(s)darr +H_2O(l)`

So at most, `1` `mol` calcium carbonate could be collected:

`"Mass of calcium carbonate"` `=` `1*molxx100.09*g*mol^-1`.

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