\(n_{H_2} = \dfrac{448}{1000.22,4} = 0,02(mol)\\ \Rightarrow n_{HCl} = 2n_{H_2} = 0,02.2 = 0,04(mol)\\ m_{muối} = m_X + m_{HCl} - m_{H_2} = 1,53 + 0,04.36,5 - 0,02.2 = 2,95\ gam\)
Ta có: \(n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
Bảo toàn nguyên tố: \(n_{Cl}=n_{HCl}=2n_{H_2}=0,04\left(mol\right)\)
\(\Rightarrow m_{Cl}=0,04\cdot35,5=1,42\left(g\right)\) \(\Rightarrow m_{muối}=1,53+1,42=2,95\left(g\right)\)