a. \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)
PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
Theo PT ta có: \(n_{O_2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
b. PTHH (2): \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Theo PT (2) ta có: \(n_{Al}=\dfrac{4}{2}.n_{O_2}=\dfrac{4}{2}.0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)
a) \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)
PTHH: \(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\uparrow\left(1\right)\)
Theo PTHH (1) : \(n_{KMnO_4}:n_{O_2}=2:1\)
\(\Rightarrow n_{O_2}=n_{KMnO_4}.\dfrac{1}{2}=0,2.\dfrac{1}{2}=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
b) Ta có \(n_{O_2}=0,1\left(mol\right)\) ( câu a )
PTHH: \(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\left(2\right)\)
Theo PTHH (2): \(n_{Al}:n_{O_2}=4:3\)
\(\Rightarrow n_{Al}=n_{O_2}.\dfrac{3}{4}=0,1.\dfrac{3}{4}=0,075\left(mol\right)\)
\(\Rightarrow m_{Al}=0,075.27=2,025\left(g\right)\)