$n_{NaOH}=0,016.0,5=0,008mol$
10ml dd cần 0,008 mol NaOH.
=> 200ml dd cần 0,16 mol NaOH.
$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
=> $n_{H_2SO_4}=0,08 mol$
$H_2SO_4.nSO_3+ nH_2O\to (n+1)H_2SO_4$
=> $n_{\text{oleum}}= \frac{0,08}{n+1}$
=> $M_{\text{oleum}}=\frac{6,58(n+1)}{0,08}= 82,25n+82,25=98+80n$
$\Leftrightarrow n= 7$
Vậy oleum $H_2SO_4.7SO_3$