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14.

`Mg+2HCl->MgCl_2+H_2`

`n_{H_2}={1,12}/{22,4}=0,05(mol)`

Theo PT: `n_{Mg}=n_{H_2}=0,05(mol)`

`->\%m_{Mg}={0,05.24}/{9,65}.100\%\approx 12,435\%`

`->\%m_{Cu}=100-12,435=87,565\%`

$\text{Mg + 2HCl → $MgCl_2$ + $H_2$↑}$

$\text{Cu + HCl → × (Cu đứng sau H)}$

$\text{Ta có : $n_{H_2}$ = $\dfrac{1,12}{22,4}$ = 0,05 (mol)}$

$\text{⇒ $m_{Mg}$ = 0,05 . 24 = 1,2 (g)}$

$\text{⇒ %Mg = $\dfrac{1,2}{9,65}$ . 100% ≈ 12,44% }$

$\text{⇒ %Cu = 100% $-$ 12,44% = 87,56%}$