Question

2HCl + CaCO3 = CaCl2 + H2O + CO2. how many mL of CO2 at 18 celsius and 715 mm Hg can be generated from 87 grams of calcium carbonate?

Answer 1

Approx. `22000*mL...`

Explanation:

We address the stoichiometric equation...

`CaCO_3(s) + 2HCl(aq) rarr CaCl_2(aq) + H_2O(l) + CO_2(g)uarr`

And thus moles of carbonate is equivalent to the moles of `CO_2(g)`..

And so `n_"calcium carbonate"` `=(87*g)/(100.09*g*mol^-1)=0.869*mol`.

And thus we need to find the volume of a `0.869*mol` quantity of carbon dioxide gas at `291.2` `K` and `715*mm*Hg` pressure...

`V=(nRT)/P=(0.869*molxx0.0821*(L*atm)/(K*mol)xx291.2*K)/((715*mm*Hg)/(760*mm*Hg*atm^-1))=??*L`

Multiply by 1000 to get an answer in `mL`...