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Answer 1 · 2017-02-15T16:14:14

Number of molecules #=6.022*10^22#

The equation of the combustion is

#C+O_2->CO_2#

If we had #12 g# of #C#, the number of molecules liberated is

#=N_A=6.022*10^23# molecules

But, we have #"0.1 g atom = 1.2 g"#

So,

the number of molecules of #CO_2#

#=(6.022*10^23)/12*1.2#

#=6.022*10^22# molecules