Here is the equation of reaction (Combustion of Carbon)

#CO_2 -> C + O_2#

**Mass of solute** #= 6g#

**Molar mass of Carbon** #= 12gmol^-1#

**Recall** #-># **Amount in moles** #= (Mass)/(Molar mass)#

#"Amount in moles" = (Mass of Carbon)/(Molar mass of carbon)#

#"Amount in moles" = 6/12#

#"Amount in moles of C" =0.5 mols#

From the stoichiometric equation we have #1 : 1 : 1# as the ratio

Since #1 mol# of #CO_2# yields #1 mol# of #C#

#:.# #0.5 mol# of #CO_2# yields #0.5mol# of #C#

Hence #"No of moles of" CO_2 = 0.5 mols#

**Also Recall** #-># **Amount in moles** #= "Volume of solution"/"Molar volume"#

#-># #"Volume of" CO_2 = "Number of moles of" CO_2 xx "Molar Volume"#

**Note that:** #"Molar Volume" = 22.5mol^-1dm^3 "at s.t.p"#

#"Volume of" CO_2 = 0.5cancel(mol) xx 22.5cancel(mol^-1)dm^3#

#"Volume of" CO_2 = (0.5 xx 22.5) dm^3#

#Volume of CO_2 = 11.25 dm^3#