\(m_{ddHCl}=D\cdot V=1,17\cdot85,4=99,918g\)
\(m_{ctHCl}=\dfrac{99,918\cdot14,6\%}{100\%}=14,59g\)
\(n_{HCl}=\dfrac{m_{HCl}}{36,5}=0,4mol\)
\(m_{ddHCl}=\dfrac{85,4}{1,17}=73\left(g\right)\\ m_{HCl}=14,6\%.73=10,658\left(g\right)\\ n_{HCl}=\dfrac{10,658}{36,5}=0,292\left(mol\right)\)