Ta có: \(n_{O_2}=\dfrac{2,4}{32}=0,075\left(mol\right)\)
PT: \(2KNO_3\underrightarrow{t^o}2KNO_2+O_2\)
Theo PT: \(n_{KNO_3\left(LT\right)}=2n_{O_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{KNO_3\left(LT\right)}=0,15.101=15,15\left(g\right)\)
Mà: H = 85%
\(\Rightarrow m_{KNO_3\left(TT\right)}=\dfrac{15,15}{85\%}\approx17,82\left(g\right)\)