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

Em tham khảo nha:

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