Nếu lấy cùng khối lượng a gam.
\(\eqalign{ & 2KMn{O_4}\,\buildrel {{t^0}} \over \longrightarrow {K_2}Mn{O_4} + Mn{O_2} + {O_2} \uparrow \cr & \,\,\,\,{a \over {158}}\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;{a \over {316}} \cr & 2KCl{O_3}\,\buildrel {{t^0}} \over \longrightarrow 2KCl + 3{O_2} \uparrow \cr & \,\,{a \over {122,5}}\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{3a} \over {245}} \cr & 2{H_2}{O_2}\,\buildrel {{t^0}} \over \longrightarrow 2{H_2}O + {O_2} \uparrow \cr & \,\,\,\,\,{a \over {34}}\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a \over {68}} \cr} \)
Ta có
\({a \over {316}} = {{a.4165} \over {316.4165}} = {{4165a} \over {1316140}}\) ứng với \({V_1}\) lít O2.
\({{3a} \over {245}} = {{3a.5372} \over {245.5372}} = {{16116a} \over {1316140}}\) ứng với V2 lít O2.
\({a \over {34}} = {{a.38710} \over {34.38710}} = {{38710a} \over {1316140}}\) ứng với V3 lít O2.
Vậy \({V_3} > {V_2} > {V_1}\).
b) Nếu lấy cùng số mol là b mol.
\(\eqalign{ & 2KMn{O_4}\,\buildrel {{t^0}} \over \longrightarrow {K_2}Mn{O_4} + Mn{O_2} + {O_2} \uparrow \cr & \,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;{n_1} = {b \over 2} \cr & 2KCl{O_3}\,\buildrel {{t^0}} \over \longrightarrow 2KCl + 3{O_2} \uparrow \cr & \,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{n_2} = {{3a} \over 2} \cr & 2{H_2}{O_2}\,\buildrel {{t^0}} \over \longrightarrow 2{H_2}O + {O_2} \uparrow \cr & \,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{n_3} = {b \over 2} \cr} \)
Ta có \({n_1} = {n_3} < {n_2} \Rightarrow {V_1} = {V_3} < {V_2}\).
