Câu 1:
\(\eqalign{ & {K_2}O + {H_2}O \to 2KOH \cr & C{O_2} + {H_2}O \to {H_2}C{O_3} \cr} \)
Câu 2:
\(\eqalign{ & {n_{BaC{l_2}}} = {{4,16} \over {208}} = 0,02mol \cr & {n_{AgN{O_3}}} = {{3,4} \over {170}} = 0,02mol. \cr & BaC{l_2} + 2AgN{O_3} \to 2AgCl + Ba{(N{O_3})_2} \cr} \)
\({n_{BaC{l_2}}}\) dư 0,01 mol, \({n_{Ba(N{O_3})}} = 0,01mol.\)
\(\eqalign{ & CM_{Ba{(N{O_3})_2}} = {{0,01} \over {0,2}} = 0,05M \cr & CM_{BaC{l_2} }= {{0,01} \over {0,2}} = 0,05M. \cr} \)
Câu 3:
\(\eqalign{ & 2KCl{O_3} \to 2KCl + 3{O_2}(xt,{t^0}) \cr & 2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}({t^0}) \cr & {{{n_{KCl{O_3}}}} \over {{n_{KMn{O_4}}}}} = {2 \over 3}:2 = {{0,66} \over 2} \cr&\Rightarrow {{{m_{KCl{O_3}}}} \over {{m_{KMn{O_4}}}}} = {{0,66.122,5} \over {2.158}} = {{81,66} \over {316}} = {1 \over {3,87}} \cr} \)
