C1
\(m_{KClO_3\left(pư\right)}=\dfrac{24,5.60}{100}=14,7\left(g\right)\)
\(n_{KClO_3\left(pư\right)}=\dfrac{14,7}{122,5}=0,12\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
0,12--------->0,12--->0,18
=> \(m_{O_2}=0,18.32=5,76\left(g\right)\)
C2: \(m_{KCl}=0,12.74,5=8,94\left(g\right)\)
Theo ĐLBTKL: \(m_{KClO_3\left(pư\right)}=m_{KCl}+m_{O_2}\)
=> \(m_{O_2}=14,7-8,94=5,76\left(g\right)\)
ta có
\(n_{KClO_3}=\dfrac{24.5}{122,5}=0,2\left(mol\right)\)
PTHH : \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
0,2 0,3 ( mol)
Mà H = 60 %
\(\Rightarrow n_{O_2}=0,3.60\%=0,18\left(mol\right)\\ \Rightarrow m_{O_2}=0,18.32=5,76\left(g\right)\)
