Em tham khảo nha:
\(\begin{array}{l}
3)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{{4,8}}{{24}} = 0,2\,mol\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,2\,mol\\
{m_{MgC{l_2}}} = 0,2 \times 95 = 19g\\
b)\\
{n_{HCl}} = 2{n_{Mg}} = 0,4\,mol\\
{C_M}HCl = \dfrac{{0,4}}{{0,4}} = 1M\\
c)\\
F{e_3}{O_4} + 4{H_2} \xrightarrow{t^0} 3Fe + 4{H_2}O\\
{n_{{H_2}}} = {n_{Mg}} = 0,2\,mol\\
{n_{F{e_3}{O_4}}} = \dfrac{{0,2}}{4} = 0,05\,mol\\
{m_{F{e_3}{O_4}}} = 0,05 \times 232 = 11,6g\\
4)\\
{n_{{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = 0,3 \times \frac{2}{3} = 0,2\,mol\\
{m_{KCl{O_3}}} = 0,2 \times 122,5 = 24,5g\\
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = 2 \times 0,3 = 0,6\,mol\\
{m_{KMn{O_4}}} = 0,6 \times 158 = 94,8g
\end{array}\)
Vậy : $KClO_3$ dùng ít hơn
