$n_{Na}=\dfrac{4,8}{23}=\dfrac{9}{115}(mol)$
Phương trình:
$2Na+2H_2O\to 2NaOH+H_2$
$\dfrac{9}{115}$ $\to$ $\dfrac{9}{115} \text{(mol)}$
$m_{NaOH\text{..tt}}=\dfrac{9}{115}.40\approx 3,13g$
$n_{H_2O}=n_{NaOH}=\dfrac{9}{115}(mol)$
Bảo toàn khối lượng:
$\to m_{\text{..tt}}=m_{Na}+m_{H_2O}$
$\to m_{\text{..tt}}=4,8+\dfrac{9}{115}.18\approx 6,21g$