\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\\ n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ Vì:\dfrac{0,3}{2}>\dfrac{0,02}{3}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=0,3-\dfrac{2}{3}.0,02=\dfrac{43}{150}\left(mol\right)\\ m_{O_2\left(dư\right)}=\dfrac{43}{150}.32=\dfrac{688}{75}\left(g\right)\)
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