\(n_{CO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
\(\Rightarrow m_{CO_2}=0,75\times44=33\left(g\right)\)
\(\Rightarrow\Sigma m_{hh}=33+16=49\left(g\right)\)
\(\%m_{CO_2}=\dfrac{33}{49}\times100\%\approx67,35\%\)
\(\%m_{SO_2}=\dfrac{16}{49}\times100\%\approx32,65\%\)