$\begin{cases}Na:x mol \\ Na_2O:y mol\\NaOH:z mol\\Na_2CO_3:t mol \end{cases}\Rightarrow \begin{cases}n_{H_2SO_4}=0,5x+y+0,5z+t \\0,5x+t=\frac{8,96}{22,4}\\\frac{2.0,5x+44t}{0,5x+t}=2.16,75\\142(0,5x+y+0,5z+t)=170,4 \end{cases}$$\Rightarrow \begin{cases}0,5x+y+0,5z+t=1,2 mol \\x=0,2 mol\\t=0,3 mol \end{cases}$
Mặt khác :
$m_{dd H_2SO_4}=\frac{1,2.98.100}{40}=294 gam\Rightarrow m_Y=m+m_{dd H_2SO_4}-m_{H_2}-m_{CO_2}$
$\Rightarrow m_Y=m+280,6$
$\Rightarrow \frac{170,4}{m+280,6}=\frac{51,449}{100}\Rightarrow m=32,1179 gam$