n Na=$\frac{26}{23}$≈1,13 mol
2Na+2H2O→2NaOH+H2↑
1,13→ 1,13 0,565 mol
-Dd A:NaOH
-Khí B:H2
mdd sau pứ=m Na+m H2O-m H2
=26+178-0,565.2 =202,87 g
m ct NaOH=1,13.40=45,2 g
⇒C% NaOH=$\frac{45,2}{202,87}$.100 ≈22,28 %
V H2(đktc)=0,565.22,4=12,656 l
--------------------Nguyễn Hoạt---------------------