Bài 1:

$n_{Zn}=\dfrac{6,5}{65}=0,1(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{H_2}=n_{Zn}=0,1(mol)\\ \Rightarrow V_{H_2}=0,1.24,79=2,479(l)\\ c,n_{ZnCl_2}=n_{Zn}=0,1(mol)\\ \Rightarrow m_{ZnCl_2}=0,1.136=13,6(g)$

Bài 2:

$n_{Mg}=\dfrac{2,4}{24}=0,1(mol)\\ a,PTHH:Mg+2HCl\to MgCl_2+H_2\\ b,n_{H_2}=n_{Mg}=0,1(mol)\\ \Rightarrow V_{H_2}=0,1.24,79=2,479(l)\\ c,n_{MgCl_2}=n_{Mg}=0,1(mol)\\ \Rightarrow m_{MgCl_2}=0,1.95=9,5(g)$

Bài 3:

$n_{H_2}=\dfrac{2,479}{24,79}=0,1(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ b,n_{Fe}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ c,n_{FeCl_2}=n_{Fe}=0,1(mol)\\ \Rightarrow m_{FeCl_2}=0,1.127=12,7(g)$

Bài 4:

$n_{H_2}=\dfrac{4,958}{24,79}=0,2(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ b,n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}(mol)\\ \Rightarrow m_{Al}=\dfrac{2}{15}.27=3,6(g)\\ c,n_{AlCl_3}=n_{Al}=\dfrac{2}{15}(mol)\\ \Rightarrow m_{AlCl_3}=\dfrac{2}{15}.133,5=17,8(g)$

Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 24,79 = 2,479l\\
c)\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,1mol\\
{m_{ZnC{l_2}}} = 0,1 \times 136 = 13,6g\\
2)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
b)\\
{n_{Mg}} = \dfrac{{2,4}}{{24}} = 0,1mol\\
{n_{{H_2}}} = {n_{Mg}} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 24,79 = 2,479l\\
c)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,1mol\\
{m_{MgC{l_2}}} = 0,1 \times 95 = 9,5g\\
3)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{2,479}}{{24,79}} = 0,1mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
c)\\
{n_{FeC{l_2}}} = {n_{{H_2}}} = 0,1mol\\
{m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g\\
4)\\
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{4,958}}{{24,79}} = 0,2mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = \frac{2}{{15}}mol\\
{m_{Al}} = \dfrac{2}{{15}} \times 27 = 3,6g\\
c)\\
{n_{AlC{l_3}}} = {n_{Al}} = \frac{2}{{15}}mol\\
{m_{AlC{l_3}}} = \dfrac{2}{{15}} \times 133,5 = 17,8g
\end{array}\)