\(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(\dfrac{2}{2}\) = \(\dfrac{2}{2}\) ( mol )
2 2 2 2 ( mol )
( Cả 2 chất đều không dư )
\(m_{H_2}=1.2=2g\)
\(m_{NaOH}=2.40=80g\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
2 2 1 ( mol )
\(m_{H_2}=1.2=2g\)
\(m_{NaOH}=2.40=80g\)