Đáp án:
a)
`Fe` + `2``HCl` → `FeCl_2` + `H_2`
b)
`m_(HCl)` = `(182,5 × 5)/(100)` = `9,125` `gam`
`n_(HCl)` = `(9,125)/(36,5)` = `0,25` `mol`
`n_(Fe)` = `(0,25 × 1)/2` = `0,125` `mol`
`m_(Fe)` = `0,125` × `56` = `7` `gam`
`n_{H_2}` = `(0,25 × 1)/2` = `0,125` `mol`
`V_{H_2}` = `0,125` × `22,4` = `2,8` `l`
`m_{H_2}` = `0,125` × `2` = `0,25` `gam`
c)
`m_{dd sau pứ}` = `7` + `182,5` - `0,25` = `189,25` `gam`
`n_{FeCl_2}` = `(0,25 × 1)/2` = `0,125` `mol`
`m_{FeCl_2}` = `0,125` × `127` = `15,875` `gam`
`C%_{FeCl_2}` = `(15,875)/(189,25)` × `100%` = `8,4%`
Giải thích các bước giải: