Question

MN LÀM dùm e bài này với ạ

Answer 1

\(n_{KClO_3}=\dfrac{73,5}{122,5}=0,6\left(mol\right)\)

\(n_{KCl}=\dfrac{33,5}{74,5}=\dfrac{67}{149}\left(mol\right)\)

PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

                a------------>a

            4KClO₃ --t^o--> 3KClO₄ + KCl

                b-------------------->0,25b

=> \(\left\{{}\begin{matrix}a+b=0,6\\a+0,25b=\dfrac{67}{149}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=0,39955\left(mol\right)\\b=0,20045\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%_{phân.hủy.theo.\left(a\right)}=\dfrac{0,39955}{0,6}.100\%=66,5917\%\\\%_{phân.hủy.theo.\left(b\right)}=\dfrac{0,20045}{0,6}.100\%=33,4083\%\end{matrix}\right.\)