$n_{Fe}$ $\text{= 16,8 : 56 = 0,3 (mol)}$$\\$$\text{Theo PTHH: 3 mol Fe phản ứng hết với 2 mol O}$$_{2}$ $\\$ $\text{Vậy: 0,3 mol Fe phản ứng hết với x mol O}$$_{2}$$\\$$\Rightarrow$ $\text{x = O$_{2}$ mol}$$\\$$\Longrightarrow$ $V_{O_{2}}$ $\text{0,2 . 22,4 = 4,48 (lít)}$
Đáp án:4,48 lít
Giải thích các bước giải:
PTHH: 3Fe + 2O2-> Fe3O4
Ta có: mFe= 16,8(g) ==> nFe=0,3(mol)
Theo PTHH, ta có: 2.nFe=3.nO2 ==>nO2=0,2(mol) ==> VO2 =4,48(l)
Vậy...
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