\(2C_nH_{2n+2}O+2Na--->2C_nH_{2n+1}ONa+H_2\)
\(n_{H_2}\left(đktc\right)=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,3\left(g\right)\)
Theo PTHH: \(n_{Na}=0,3\left(mol\right)\)
\(\Rightarrow m_{Na}=6,9\left(g\right)\)
Ap dung ĐLBTKL, ta có:
\(\Rightarrow m=20,4+0,3-6,9=13,8\left(g\right)\)