a) 2KClO_3 $$\xrightarrow{t^o}$$ 2KCl+3O_2 BTKL: n_{O_2\ t t}={50-38}/{32}=0,375(mol) Theo pt: n_{KClO_3\ t t}=2/3n_{O_2\ t t}=0,25(mol) H={0,25.122,5}/{50}.100\%=61,25\% b) n_{KCl\ t t}=n_{KClO_3\ t t}=0,25(mol) m_{KCl\ t t}=0,25.74,5=18,625(g) m_{KClO_3\ du]=38-18,625=19,375(g)
