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#
