\(V_{C_2H_5OH\left(\text{nguyên chất}\right)}=16.71,875\%=11,5\left(ml\right)\\ \rightarrow m_{C_2H_5OH\left(\text{nguyên chất}\right)}=11,5.0,8=9,2\left(g\right)\\ \rightarrow n_{C_2H_5OH\left(\text{nguyên chất}\right)}=\dfrac{9,2}{46}=0,2\left(mol\right)\)
PTHH: 2C2H5OH + 2K ---> 2C2H5OK + H2
0,2 0,1
=> VH2 = 0,1.22,4 = 2,24 (l)