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")#