Ta có : \(\text{mH2SO4=19,6%.150=29,4 gam}\)\(\rightarrow\)nH2SO4=0,3 mol
\(\rightarrow\) mH2O=150-29,4=120,6 gam \(\rightarrow\) nH2O=\(\frac{120,6}{18}\)=6,7 mol
Phản ứng:
2Na + H2SO4 \(\rightarrow\) Na2SO4 + H2
Mg + H2SO4 \(\rightarrow\) MgSO4 + H2
2Na + 2H2O \(\rightarrow\) 2NaOH + H2
Ta có: nH2=nH2SO4 + \(\frac{1}{2}\)nH2O=0,3+\(\frac{6,7}{2}\)=3,65 mol
\(\rightarrow\) V H2=3,65.22,4=81,76 lít