`M+2HCl->MCl_2+H_2`
Ta có
`n_(M)=\frac{10,92}{M}`
Theo PT
`n_(MCl_2)=\frac{10,92}{M}(mol)`
`=>m_(MCl_2)=\frac{10,92M+775,32}{M}(g)`
`n_(HCl)=\frac{21,84}{M} (mol)`
`m_(dd HCl)=\frac{7971,6}{M}(g)`
`n_(H_2)=\frac{10,92}{M}(mol)`
`=>m_(H_2)=\frac{21,84}{M} (g)`
`=>m_(dd)=10,92+\frac{7971,6}{M}-\frac{21,84}{M}=10,92+\frac{7949,76}{M}`
Lại có
`C%_(MCl_2)=16,2%`
`=\frac{\frac{10,92M+775,32}{M}}{10,92+\frac{7949,76}{M}}=\frac{16,2}{100}`
`=>M=56`
`=>M` là `Fe`
`Fe+2HCl->FeCl_2+H_2`
`n_(H_2)=\frac{10,92}{56}=0,195(mol)`
`=>V_(H_2)=0,195.22,4=4,368(l)`
