Co ai khong giup to voi to cần gấp ngay bay gio

Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\,2K + 2{H_2}O \to 2KOH + {H_2}\\
b)\,S{O_3} + {H_2}O \to {H_2}S{O_4}\\
c)\,3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
d)\,Zn + 2HCl \to ZnC{l_2} + {H_2}\\
e)\,2{H_2} + {O_2} \xrightarrow{t^0} 2{H_2}O\\
f)\,4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
2)\\
1.\,\\
Natri\,clorua:NaCl\\
Axit\,sunfuric:{H_2}S{O_4}\\
\text{ Nhôm hidroxit } :Al{(OH)_3}\\
Kali\,hidrocacbonat:KHC{O_3}:\\
\text { Đồng II sunphat }:CuS{O_4}\\
2)\\
Ca{(OH)_3} \to Ca{(OH)_2}\\
MgHC{O_3} \to Mg{(HC{O_3})_2}\\
{H_3}N{O_3} \to HN{O_3}\\
3)\\
1.\,{n_{NaCl}} = 0,05 \times 1 = 0,05\,mol\\
{m_{NaCl}} = 0,05 \times 58,5 = 2,925g\\
2.\\
{m_{{\rm{dd}}NaOH}} = 20 + 380 = 400g\\
{C_\% }NaOH = \dfrac{{20}}{{400}} \times 100\%  = 5\% \\
4)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{22,4}}{{22,4}} = 1\,mol\\
{n_{HCl}} = 2{n_{{H_2}}} = 2\,mol\\
{C_M}HCl = \dfrac{2}{{0,05}} = 40M
\end{array}\)