Gọi `x = n_(Na) ; y = n_(Fe) `
`n_(H_2) = 6.72 / 22.4 = 0.3 ( mol )`
`PTHH : Na + HCl -> NaCl + 1/2 H_2 ↑ `
` Fe + 2HCl -> FeCl_2 + H_2`
Ta có hệ pt : $\left \{ {{23x+56y=14.8} \atop {x/2+y=0.3}} \right.$ `=>`$\left \{ {{x = 0.4} \atop {y=0.1}} \right.$
`-> %_(Na) =`$\dfrac{0.4×23}{14.8}$`×100% = 62.16%`
`-> %_(Fe) = 100% - 62.16% = 37.84%`