Question #4658d

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Answer 1 · 2017-12-18T08:50:20

$= 4.83 g F e_{3} O_{4}$ $=4.83gFe_3O_4$

Find the molar mass of the involved substances
$O_{2} = \frac{32 g}{m o l}$ $O_2=(32g)/(mol)$
$F e_{3} O_{4} = \frac{232 g}{m o l}$ $Fe_3O_4=(232g)/(mol)$
Given the $m O_{2}$ $mO_2$ , convert it to $n O_{2}$ $nO_2$ through molar conversion with the relationship provided in the molar mass of $O_{2}$ $O_2$ .
$=2.00cancel(gO_2)xx(1molO_2)/(32cancel(gO_2))$
$=0.0625molO_2$
Now, find the $nFe_3O_4$ . Refer and use the balanced equation as provided for the mole ratio; i.e.,
$=0.0625cancel(molO_2)xx(1molFe_3O_4)/(3cancel(molO_2))$
$=0.02083molFe_3O_4$
Finally, find the $mFe_3O_4$ through the relationship obtainable from the molar mass of the compound to convert mole to mass; thus,
$=0.02083cancel(molFe_3O_4)xx(232gFe_3O_4)/(1cancel(molFe_3O_4))$
$=4.83gFe_3O_4$