\(n_{KClO_3}=\dfrac{2,5}{122,5}=\dfrac{1}{49}\left(mol\right)\\ PTHH:2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{O_2}=\dfrac{\dfrac{3.1}{49}}{2}=\dfrac{3}{98}\left(mol\right)\\ V_{O_2\left(đktc\right)}=\dfrac{3}{98}.22,4=\dfrac{24}{35}\left(l\right)\\ C1:n_{KClO_3}=n_{KClO_3}=\dfrac{1}{49}\left(mol\right)\Rightarrow m_{KCl}=74,5.\dfrac{1}{49}=\dfrac{149}{98}\left(g\right)\\ C2:m_{KCl}=m_{KClO_3}-m_{O_2}=2,5-\dfrac{3}{98}.32=\dfrac{149}{98}\left(g\right)\)
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