\(PTHH:C_2H_5OH+Na\rightarrow C_2H_5ONa+\frac{1}{2}H_2\)
Ta có:
\(n_{H2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
\(\Rightarrow n_{C2H5OH}=0,1\left(mol\right)\Rightarrow m_{C2H5OH}=0,1.46=4,6\left(g\right)\)
C2H5OH+Na->C2H5ONa+H2
nH2=1,12\22,4=0,05 mol
=>mC2H5OH=0,05.46=2,3g