\(a,PTHH:2KClO_3\xrightarrow{t^o}2KCl+3O_2\\ b,n_{KClO_3(thực tế)}=\dfrac{36,75}{122,5}=0,3(mol)\\ n_{O_2(phản ứng)}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{KClO_3(phản ứng)}=\dfrac{2}{3}n_{O_2}=0,2(mol)\\ \Rightarrow H\%=\dfrac{0,2}{0,3}.100\%=66,67\%\\ c,n_{KCl}=n_{KClO_3(phản ứng)}=0,2(mol)\\ \Rightarrow m_{KCl}=0,2.74,5=14,9(g)\)