\(n_{O_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)
\(0.2..................................................0.1\)
\(m_{KMnO_4}=0.2\cdot158=31.6\left(g\right)\)
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Câu 1 :
\(n_{Fe}=\dfrac{0.56}{56}=0.01\left(mol\right)\)
\(n_{O_2}=\dfrac{16}{32}=0.5\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(n_{Fe}:3< n_{O_2}:2\Rightarrow O_2dư\)
\(n_{Fe_3O_4}=\dfrac{0.01}{3}=\dfrac{1}{300}\left(mol\right)\)
\(m=\dfrac{1}{300}\cdot232=0.774\left(g\right)\)
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Câu 14 :
\(n_{NaOH}=0.4\cdot6=2.4\left(mol\right)\)
Câu 15 : C
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