When calcium carbonate is reacted with hydrochloric acid, a certain amount of `"CO"_2` evolves. If `"0.250 mols"` of `"CaCO"_3` was reacted, what volume of `"CO"_2` evolves at 100% yield?

`V_(CO_2(g)) = "5.60 L"`

Well, you'll have to identify the phases of each substance before doing anything else. This reaction is doable at room temperature, so the only gas is `"CO"_2`:

`"CaCO"_3(s) + 2"HCl"(aq) -> "CaCl"_2(aq) + cancel("H"_2"CO"_3(aq))`
`cancel("H"_2"CO"_3(aq)) -> "H"_2"O"(l) + "CO"_2(g)`
`ul(" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ")`
`color(green)("CaCO"_3(s) + 2"HCl"(aq) -> "CaCl"_2(aq) + "H"_2"O"(l) + "CO"_2(g))`

`"CaCO"_3(s)` is generally poorly soluble in water, so it's a solid. But by adding a strong acid, it reacts to dissolve `"Ca"^(2+)` into solution with the `"Cl"^(-)` from the `"HCl"`, which is generally provided as an aqueous solution.

The water is a pure liquid that comes from the carbonic acid that decomposes, since `"CO"_2` is poorly soluble in water (as are many gases).

Knowing that `"1 mol"` of `"CaCO"_3(s)` gives you `"1 mol"` of `"CO"_2(g)` from the coefficients in the reaction, `"0.250 mols"` of `"CaCO"_3(s)` gives you `"0.250 mols"` of `"CO"_2(g)`.

Therefore, the volume of ideal gas made at STP, which we define as `0^@ "C"` and `"1 atm"`, is based on the ideal gas law:

`PV = nRT`

• `P` is pressure in `"atm"`.
• `V` is volume in `"L"`.
• `n` is the mols of ideal gas.
• `R = "0.082057 L"cdot"atm/mol"cdot"K"` is the universal gas constant.
• `T` is the temperature in `"K"`.

Thus,

`color(blue)(V) = (nRT)/P`

`= (0.250 cancel("mols CO"_2) cdot "0.082057 L"cdotcancel"atm/mol"cdotcancel"K" cdot 273.15 cancel"K")/(cancel"1 atm")`

`=` `color(blue)("5.60 L")`