`#AkaShi`
PTHH:
`2Fe + 3Cl_2` $\xrightarrow{t^o}$ `2FeCl_3`
`Cu + Cl_2` $\xrightarrow{t^o}$ `CuCl_2`
`Fe + 2HCl -> FeCl_2 + H_2↑`
Gọi `x;y` lần lượt là số mol của `Fe` và `Cu`
Theo đề ta có:
`n_{FeCl_2}=n_{Fe}=x (mol)`
Mà `n_{FeCl_2}=(m_{FeCl_2})/(M_{FeCl_2})=(25,4)/(56+71)=0,2 (mol)`
Ta lại có:
`m_{hh}=59,5 (g)`
`->162,5x+135y=59,5`
`->135y=59,5-(162,5xx0,20`
`->135y=27`
`->y=0,2`
`->n_{Cu}=m_{CuCl_2}=y=0,2 (mol)`
`->m_{Cu}=n_{Cu}xxM_{Cu}=0,2xx64=12,8 (g)`
`->n_{Fe}=x=0,2 (mol)`
`->m_{Fe}=n_{Fe}xxM_{Fe}=0,2xx56=11,2 (g)`
`->%m_{Cu}=(12,8)/(12,8+11,2)xx100%=53,33%`
`->%m_{Fe}(11,2)/(11,2+12,8)xx100%=46,67%`
`\text{________________________________________________}`
`b)`
Theo pt: `Fe + 2HCl -> FeCl_2 + H_2↑`
`->n_{HCl}=2xxn_{Fe}=0,2xx2=0,4 (mol)`
`->m_{HCl}=n_{HCl}xxM_{HCl}=0,4xx36,5=14,6 (g)`
`->m_{dd\ HCl}=14,6:10%=146 (g)`
`->V_{HCl}=(146)/(0,9091)=160,6ml`
