Ta có: \(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)
PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
____0,5______________0,5____0,25 (mol)
Ta có: m dd sau pư = 11,5 + 200 - 0,25.2 = 211 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,5.40}{211}.100\%\approx9,48\%\)
`n_{Na} = (11,5)/(23) = 0,5 (mol)`
PTHH: `2Na + 2H_2O -> 2NaOH + H_2`
Theo PT: `n_{H_2} = 1/2 n_{Na} = 0,25 (mol)`
`=> m_{dd} = 11,5 + 200 - 0,25.2 = 211 (g)`
Theo PT: `n_{NaOH} = n_{Na} = 0,5 (mol)`
`=> C\%_{NaOH} = (0,5.40)/(211) .100\% = 9,48\%`
\(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{NaOH}=n_{Na}=0,5\left(mol\right)\\ m_{NaOH}=0,5.40=20\left(g\right)\\ n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{ddNaOH}=11,5+200-0,25.2=211\left(g\right)\\ C\%_{ddNaOH}=\dfrac{20}{211}.100\approx9,479\%\)
