Bài 1:
a) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
Theo PTHH: \(n_{Fe}:n_{O_2}=3:2\)
\(\Rightarrow n_{O_2}=n_{Fe}.\dfrac{2}{3}=0,1.\dfrac{2}{3}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow V_{O_2}=\dfrac{1}{15}.22,4=1,5\left(l\right)\)
b) Theo PTHH: \(n_{Fe}:n_{Fe_3O_4}=3:1\)
\(\Rightarrow n_{Fe_3O_4}=n_{Fe}.\dfrac{1}{3}=0,1.\dfrac{1}{3}=\dfrac{1}{30}\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232=7,7\left(g\right)\)
\(n_{KClO_3}=\dfrac{a}{122,5}\left(mol\right)\)
\(n_{KMnO_4}=\dfrac{b}{158}\left(mol\right)\)
PTHH: \(2KClO_3\underrightarrow{t^0}2KCl+3O_2\uparrow\left(1\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\uparrow\left(2\right)\)
Từ phương trình (1): \(n_{O_2}=\dfrac{3}{2}.n_{KClO_3}=\dfrac{3}{2}.\dfrac{a}{122,5}=\dfrac{3a}{245}\left(mol\right)\)
Từ phương trình (2): \(n_{O_2}=\dfrac{1}{2}.n_{KMnO_4}=\dfrac{1}{2}.\dfrac{b}{158}=\dfrac{b}{316}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(1\right)}=\dfrac{3a}{245}.32=\dfrac{96a}{245}\left(g\right)\\m_{O_2\left(2\right)}=\dfrac{b}{316}.32=\dfrac{8b}{79}\left(g\right)\end{matrix}\right.\)
Theo bài khối lượng thu được ở 2 phản ứng bằng nhau:
\(\Rightarrow\dfrac{96a}{245}=\dfrac{8b}{79}\Rightarrow96a.79=8b.245\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{245}{948}\)
