\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ n_{O_2}=\dfrac{3}{2}n_{KClO_3}=\dfrac{3}{2}.\dfrac{a}{122.5}=\dfrac{3a}{245}\left(mol\right)\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=\dfrac{1}{2}.\dfrac{b}{158}=\dfrac{b}{316}\left(mol\right)\)
Suy ra :
\(\dfrac{3a}{245}=\dfrac{b}{316}\Leftrightarrow\dfrac{a}{b}=\dfrac{245}{3.316}=\dfrac{245}{948}\)
$M + 2AgNO_3 \to M(NO_3)_2 + 2Ag$
Theo PTHH :
$n_{M(NO_3)_2} = \dfrac{1}{2}n_{AgNO_3} = 0,1(mol)$
$M_{M(NO_3)_2} = M + 62.2 = \dfrac{18,8}{0,1} = 188\Rightarrow M = 64(Cu)$
Vậy kim loại M là Cu
