Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{{7,2}}{{24}} = 0,3mol\\
{n_{{H_2}}} = {n_{Mg}} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
{n_{HCl}} = 2{n_{Mg}} = 0,6mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,6}}{{0,2}} = 3M\\
c)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,3mol\\
{m_{MgC{l_2}}} = 0,3 \times 95 = 28,5g\\
2)C\\
2KMn{O_4} + 16HCl \to 5C{l_2} + 2KCl + 2MnC{l_2} + 8{H_2}O\\
{n_{HCl}} = \dfrac{{29,2}}{{36,5}} = 0,8mol\\
{n_{C{l_2}}} = \dfrac{5}{{16}}{n_{HCl}} = 0,25mol\\
{V_{C{l_2}}} = 0,25 \times 22,4 = 5,6l\\
3)\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
{n_{FeC{l_3}}} = \dfrac{{32,5}}{{162,5}} = 0,2mol\\
{n_{C{l_2}}} = \frac{3}{2}{n_{FeC{l_3}}} = 0,3mol\\
{V_{C{l_2}}} = 0,3 \times 22,4 = 6,72l\\
4)\\
2NaI + Mn{O_2} + 2{H_2}S{O_4} \to {I_2} + MnS{O_4} + N{a_2}S{O_4} + 2{H_2}O\\
{n_{Mn{O_2}}} = \dfrac{{17,4}}{{87}} = 0,2mol\\
{n_{{I_2}}} = {n_{Mn{O_2}}} = 0,2mol\\
{m_{{I_2}}} = 0,2 \times 254 = 50,8g
\end{array}\)