\(M_A=d_{A/H_2}\cdot M_{H_2}=37,25\cdot2=74,5\left(g/mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_K=74,5\cdot52,35\%=39\left(g\right)\\m_{Cl}=74,5-39=35,5\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_K=\dfrac{39}{39}=1\left(mol\right)\\n_{Cl}=\dfrac{35,5}{35,5}=1\left(mol\right)\end{matrix}\right.\\ \Rightarrow CTHH_A:KCl\)