a. PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b. \(n_{Fe}=\dfrac{16}{56}=0,28\left(mol\right)\)
Theo PT ta có: \(n_{Fe_3O_4}=\dfrac{0,28.1}{3}=0,09\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=n.M=0,09.232=20,88\left(g\right)\)
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a) 3Fe + 2O2 \(\underrightarrow{to}\) Fe3O4
\(n_{Fe}=\dfrac{16}{56}=\dfrac{2}{7}\left(mol\right)\)
b) Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{3}\times\dfrac{2}{7}=\dfrac{2}{21}\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=\dfrac{2}{21}\times232=22,095\left(g\right)\)
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