a) \(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)
\(m_{H_2SO_4}=\dfrac{150.9,8}{100}=14,7\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
PTHH: 2Na + H2SO4 --> Na2SO4 + H2
Xét tỉ lệ: \(\dfrac{0,5}{2}>\dfrac{0,15}{1}\) => Na dư, H2SO4 hết
PTHH: 2Na + H2SO4 --> Na2SO4 + H2
0,3<---0,15----->0,15--->0,15
2Na + 2H2O --> 2NaOH + H2
0,2-------------->0,2---->0,1
\(n_{H_2}=0,15+0,1=0,25\left(mol\right)\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)
b)
\(\left\{{}\begin{matrix}m_{Na_2SO_4}=0,15.142=21,3\left(g\right)\\m_{NaOH}=0,2.40=8\left(g\right)\end{matrix}\right.\)
mdd sau pư = 11,5 + 150 - 0,25.2 = 161 (g)
\(\left\{{}\begin{matrix}C\%_{Na_2SO_4}=\dfrac{21,3}{161}.100\%=13,23\%\\C\%_{NaOH}=\dfrac{8}{161}.100\%=4,97\%\end{matrix}\right.\)
