Question

Answer 1

\(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\\ pthh:2KClO_3\xrightarrow[xtMnO_2]{t^o}2KCl+3O_2\) 
              0,2                                   0,3 
\(V_{O_2}=0,3.22,4=6,72\left(l\right)\)
\(n_C=\dfrac{1,2}{12}=0,1\left(mol\right)\\ pthh:C+O_2\underrightarrow{t^o}CO_2\) 
  \(LTL:\dfrac{0,1}{1}< \dfrac{0,3}{1}\) 
=> mẩu than cháy hết 
\(n_{O_2\left(p\text{ư}\right)}=n_C=0,1\left(mol\right)\\ V_{O_2\left(D\right)}=\left(0,3-0,1\right).22,4=4,48\left(l\right)\\ n_{CO_2}=n_C=0,1\left(mol\right)\\ V_{CO_2}=0,1.22,4=2,24\left(l\right)\\ V_{kh\text{í}}=2,24+4,48=6,72\left(l\right)\)

Answer 2

nC=1,212=0,1(mol)pthh:C