Bài 13:
a, Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}+\dfrac{5}{4}n_P=0,35\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,35.22,4=7,84\left(l\right)\)
b, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4\left(LT\right)}=2n_{O_2}=0,7\left(mol\right)\)
Mà: H% = 90%
\(\Rightarrow n_{KMnO_4\left(TT\right)}=\dfrac{0,7}{90\%}=\dfrac{7}{9}\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=\dfrac{7}{9}.158\approx122,89\left(g\right)\)
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