\(a) n_{HCl} = \dfrac{50.1,194.38\%}{36,5} = 0,6215(mol)\\ C_{M_{HCl}} = \dfrac{0,6215}{0,05} = 12,43M\\ b) Ca(OH)_2 + 2HCl \to CaCl_2 + 2H_2O\\ n_{Ca(OH)_2} = \dfrac{1}{2}n_{HCl} = 0,31075(mol)\\ m_{dd\ Ca(OH)_2} = \dfrac{0,31075.74}{25\%} = 91,982(gam)\)