Đáp án + Giải thích các bước giải:

$Mg+2HCl\to MgCl_2+H_2$

Theo PT: $n_{Mg}=n_{H_2}=\frac{5,6}{22,4}=0,25(mol)$

$\to \%m_{Mg}=\frac{0,25.24}{12,4}.100\%\approx 48,39\%$

$\to \%m_{Cu}=100-48,39=51,61\%$

Theo PT: $n_{MgCl_2}=n_{Mg}=0,25(mol);n_{HCl(pứ)}=2n_{Mg}=0,5(mol)$

Ta có $m_{HCl(dư)}=120.18,25\%-0,5.36,5=3,65(g)$

Ta có $m_{dd\,spu}=12,4+120-0,25.2=131,9(g)$

$\to C\%_{HCl(dư)}=\frac{3,65}{131,9}.100\%\approx 2,77\%$

$\to C\%_{MgCl_2}=\frac{0,25.95}{131,9}.100\%\approx 18,01\%$

Em tham khảo nha :

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,25mol\\
{m_{Mg}} = 0,25 \times 24 = 6g\\
{m_{Cu}} = 12,4 - 6 = 6,4g\\
\% Mg = \dfrac{6}{{12,4}} \times 100\%  = 48,39\% \\
\% Cu = 100 - 48,39 = 51,61\% \\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,25mol\\
{m_{MgC{l_2}}} = 0,25 \times 95 = 23,75g\\
{n_{HCl}} = \dfrac{{120 \times 18,25\% }}{{36,5}} = 0,6mol\\
{n_{HC{l_d}}} = 0,6 - 0,25 \times 2 = 0,1mol\\
{m_{HC{l_d}}} = 0,1 \times 36,5 = 3,65g\\
C{\% _{HCl}} = \dfrac{{3,65}}{{6 + 120 - 0,5}} \times 100\%  = 2,9\% \\
C{\% _{MgC{l_2}}} = \dfrac{{23,75}}{{6 + 120 - 0,5}} \times 100\%  = 18,9\% 
\end{array}\)