`n_{Fe}=\frac{11,2}{56}=0,2(mol)`
`a)` `Fe+2HCl\to FeCl_2+H_2`
`n_{H_2}=n_{Fe}=0,2(mol)`
`=> V_{H_2}=0,2.22,4=4,48(l)`
`b)` `n_{HCl}=2n_{H_2}=0,4(mol)`
`=> C%_{HCl}=\frac{0,4.36,5.100%}{200}=7,3%`
`c)` BTKL:
`m_{\text{ dd spu}}=m_{Fe}+m_{\text{dd} HCl}-m_{H_2}`
`=> m_{\text{dd spu}}=11,2+200-0,4=210,8g`
`=> C%_{FeCl_2}=\frac{0,2.127.100%}{210,8}\approx 12,05%`