2Na + 2H2O → 2NaOH + H2↑
\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
Theo PT: \(n_{NaOH}=n_{Na}=0,1\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,1\times40=4\left(g\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}\times0,1=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,05\times2=0,1\left(g\right)\)
\(m_{ddNaOH}=m_{Na}+m_{H_2O}-m_{H_2}=2,3+50-0,1=52,2\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{4}{52,2}\times100\%=7,66\%\)
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