a/ \(2KClO_3\left(1,6\right)\rightarrow2KCl\left(1,6\right)+3O_2\left(2,4\right)\)
\(n_{O_2}=\dfrac{53,76}{22,4}=2,4\left(mol\right)\)
\(\Rightarrow m_{KClO_3\left(pứ\right)}=1,6.122,5=196\left(g\right)\)
\(\Rightarrow m_{KCl}=1,6.74,5=119,2\left(g\right)\)
\(\Rightarrow m-196+119,2=168,2\)
\(\Leftrightarrow m=245\left(g\right)\)
\(\Rightarrow\%KClO_3\left(pứ\right)=\dfrac{196}{245}.100\%=80\%\)
b/ \(2KMnO_4\left(4,8\right)\rightarrow K_2MnO_4+MnO_2+O_2\left(2,4\right)\)
\(\Rightarrow m_{KMnO_4}=4,8.158=758,4\left(g\right)\)
Khối lượng thuốc tím cần là: \(\dfrac{758,4}{90\%}\approx842,67\left(g\right)\)