\(2KCl{O_3}\buildrel {{t^o}} \over\longrightarrow 2KCl + 3{O_2} \uparrow \)
2 mol 3 mol
\(\dfrac{{a}}{{122,5}}mol\) \(\dfrac{{3a}}{{{2 \times 122,5}}}mol\)
\(2KMn{O_4}\buildrel {{t^o}} \over\longrightarrow {K_2}Mn{O_2} + {O_2} \uparrow + Mn{O_2}\)
2 mol 1 mol
\(\dfrac{{b}}{{158}}mol\) \(\dfrac{{b}}{{{2 \times 158}}}mol\)
Muốn được cùng một lượng oxi: \(\dfrac{{3a}}{{{2 \times 122,5}}}\) = \(\dfrac{{b}}{{{2 \times 158}}}\)
Rút ra tỷ lệ:
\(\dfrac{a}{b} = \dfrac{245}{{948}} = \dfrac{7}{27,0875} \)
