\(n_{KMnO_4}=1\left(mol\right)\\ a,2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\uparrow\\ a,n_{O_2}=\dfrac{n_{KMnO_4}}{2}=\dfrac{1}{2}=0,5\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,5.22,5=11,2\left(l\right)\\ \Rightarrow V_{kk}=5.V_{O_2\left(đktc\right)}=5.11,2=56\left(l\right)\\ b,2Mg+O_2\rightarrow\left(t^o\right)2MgO\\ n_{MgO}=2.n_{O_2}=2.0,5=1\left(mol\right)\\ \Rightarrow m_{MgO}=1.40=40\left(g\right)\)
a. PTHH: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
- Theo PTHH ta có:
\(n_O=n_{KMnO_4}\cdot\dfrac{1}{2}=1\cdot\dfrac{1}{2}=0,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)
- Do thể tích Oxi chiếm 1/5 thể tích không khí nên:
\(V_{KK}=11,2.5=56\left(l\right)\)