Cho 11,2g iron + V(lít) O2 đkc -> thu được m(g) oxit sắt từ Fe3O4

Đáp án:

$\rm n_{Fe}=\dfrac{11,2}{56}=0,2\ (mol)$

Phản ứng xảy ra:

$\rm 3Fe+2O_2\xrightarrow{t^°}Fe_3O_4$

Ta có:

$\rm n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{15}\ (mol)\\n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{15}\ (mol)$

$\rm\to V_{O_2}=\dfrac{2}{15}×24,79≈3\ (L)\\\to m_{Fe_3O_4}=\dfrac{1}{15}×232≈15,5\ (g)$

`n_{Fe}={11,2}/{56}=0,2(mol)`

`3Fe+2O_2`  $\xrightarrow{t^o}$  `Fe_3O_4`

`0,2->2/{15}->1/{15}(mol)`

`V=2/{15}.24,79\approx 3,3053(l)`

`m=1/{15}.232\approx 15,47(g)`