Đáp án + Giải thích các bước giải:

\(\begin{array} {l} a)\\ n_{Mg}=\dfrac{4,8}{24}=0,2(mol)\\ n_{HCl}=\dfrac{182,5.10\%}{36,5}=0,5(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ n_{Mg}<\dfrac{n_{HCl}}{2}\to HCl\text{ dư}\\ \text{Theo PT: }n_{HCl(pứ)}=2n_{Mg}=0,4(mol)\\ \to n_{HCl(dư)}=(0,5-0,4).36,5=3,65(g)\\ b)\\ \text{Theo PT: }n_{H_2}=n_{Mg}=0,2(mol)\\ \to V_{H_2(đkc)}=0,2.24,79=4,958(l)\\ \to V_{H_2(đkc)(tt)}=4,958.85\%=4,2143(l)\\ c)\\ \text{Theo PT: }n_{MgCl_2}=n_{Mg}=0,2(mol)\\ \to m_{MgCl_2(tt)}=0,2.95.85\%=16,15(g) \end{array}\)

Em tham khảo nha :

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{{4,8}}{{24}} = 0,2mol\\
{n_{HCl}} = \dfrac{{182,5 \times 10\% }}{{36,5}} = 0,5mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,5}}{2} \Rightarrow HCl\text{ dư}\\
{n_{HC{l_d}}} = 0,5 - 0,2 \times 2 = 0,1mol\\
{m_{HC{l_d}}} = 0,1 \times 36,5 = 3,65g\\
b)\\
{n_{{H_2}}} = {n_{Mg}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 24,79 = 4,958l\\
c)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,2mol\\
{m_{MgC{l_2}}} = 0,2 \times 95 = 19g\\
{m_{MgC{l_2}td}} = \dfrac{{19 \times 85}}{{100}} = 16,15g
\end{array}\)