Đáp án đúng: D
Giải chi tiết:\(\begin{array}{l}{m_{X\,(ct)}} = \frac{{70,8.15}}{{100}} = 10,62\,gam\\BTKL:{m_X} + {m_{HCl}} = {m_{muoi}} \Rightarrow {m_{HCl}} = 17,19 - 10,62 = 6,57\,gam\\\Rightarrow {n_{HCl}} = 0,18\,mol\\\,\,\,\,\,\,\,\,\,1X\,\,\, + \,\,\,1HCl \to muoi\\\,\,\,\,\,\,0,18\, \leftarrow \,\,\,0,18\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(mol)\\\Rightarrow {M_X} = \frac{{10,62}}{{0,18}} = 59 \Rightarrow X:{C_3}{H_9}N.\end{array}\)
Vì X là amin bậc hai nên X là C2H5NHCH3 (etylmetylamin)
Đáp án D