PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)
Ta có: \(n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,3\cdot40=12\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Na}+m_{H_2O}-m_{H_2}=126,6\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{12}{126,6}\cdot100\%\approx9,48\%\)
Đúng 2
Bình luận (0)
