$a) n_{Na} = \dfrac{46}{23} = 2(mol)$
$2Na + 2H_2O \to 2NaOH + H_2$
Theo PTHH :
$n_{H_2} = \dfrac{1}{2}n_{Na} = 1(mol)$
$V = 1.22,4 = 22,4(lít)$
b) $m_{dd\ sau\ pư} = 46 + 100 - 1.2 = 144(gam)$
$C\%_{NaOH} = \dfrac{2.40}{144}.100\% = 55,56\%$
Bài 1:
a, Ta có: \(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)
PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
_____2______________2______1 (mol)
\(\Rightarrow V_{H_2}=1.22,4=22,4\left(l\right)\)
b, Có: m dd sau pư = mNa + mH2O - mH2 = 46 + 1000 - 1.2 = 1044 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{2.40}{1044}.100\%\approx7,66\%\)
