Câu 1:
\(n_C=\dfrac{1,5}{12}=0,125\left(mol\right)\)
PTHH: C + O2 --to--> CO2
0,125->0,125
=> VO2 = 0,125.22,4 = 2,8 (l)
=> Vkk = 2,8.5 = 14 (l)
Câu 2:
\(n_{KClO_3}=\dfrac{6,125}{122,5}=0,05\left(mol\right)\)
PTHH: 2KClO3 --to,MnO2--> 2KCl + 3O2
0,05----------------------->0,075
=> \(V_{O_2}=0,075.22,4=1,68\left(l\right)\)