Bài 11:
Ta có: \(\left\{{}\begin{matrix}n_P=\dfrac{5}{31}\left(mol\right)\\n_{O_2}=\dfrac{2,8.20\%}{22,4}=0,025\left(mol\right)\end{matrix}\right.\)
PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)
Xét tỉ lệ: \(\dfrac{\dfrac{5}{31}}{4}>\dfrac{0,025}{5}\) => P dư
Theo PT: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,01\left(mol\right)\)
=> \(m_{P_2O_5}=0,01.142=1,42\left(g\right)\)
Bài 12:
Ta có: \(\left\{{}\begin{matrix}n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right)\\n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\end{matrix}\right.\)
PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,05\left(mol\right)\)
PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
Xét tỉ lệ: \(\dfrac{0,1}{3}>\dfrac{0,05}{2}\) => Fe dư, lượng O2 không đủ
