\(n_{Na}=\dfrac{6,9}{23}=0,3mol\)
\(n_{H_2O}=\dfrac{150}{18}=\dfrac{25}{3}mol\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(0,3\) \(\dfrac{25}{3}\) 0 0
0,3 0,3 0,3 0,15
\(m_{NaOH}=0,3\cdot40=12g\)
\(m_{H_2}=0,15\cdot2=0,3g\)
\(m_{ddNaOH}=6,9+150-0,3=156,6g\)
\(C\%=\dfrac{12}{156,6}\cdot100\%=7,66\%\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\)
\(nNa=\dfrac{6,9}{23}=0,3\left(mol\right)\)
\(nH_2O=\dfrac{150}{18}=8,3\left(mol\right)\)
Xét tỉ lê :
\(\dfrac{0,3}{2}< \dfrac{8,3}{2}\)
H2O dư , tính số mol dd theo số mol của Na
\(\Rightarrow nNaOH=nNa=0,3\left(mol\right)\)
\(\Rightarrow mNaOH=0,3.40=12\left(g\right)\)
\(C\%=\dfrac{12}{6,9+150}.100\%=7,64\%\)
