`n_{Mg}=\frac{3,6}{24}=0,15(mol)`
`Mg+2HCl\to MgCl_2+H_2`
`n_{H_2}=n_{Mg}=0,15(mol)`
`=> V_{H_2}=0,15.22,4=3,36(l)`
`n_{HCl}=2n_{H_2}=0,3(mol)`
`V_{HCl}=\frac{0,3}{1}=0,3(l)`
`m_{\text{dd HCl}}=d.V=300.1,2=360g`
BTKL:
`m_{\text{dd spu}}=m_{Mg}+m_{\text{dd} HCl}-m_{H_2}`
`=> m_{\text{dd spu}}=3,6+360-0,15.2=363,3g`
`C%_{MgCl_2}=\frac{95.0,15.100%}{363,3}\approx 3,92%`