Question

4 fe+302_2 fe2o3 5.6 grams of fe reacts 3 moles of o2 find the mass of ferrous oxide​

Answer 1

Answer:

4fe+302_2 fe2o3 5.6 grams of fe reacts 3 moles of o2 find the mass of ferrous oxide. 1. See answer. Add ans

Answer 2

Given:

$4Fe + 3O_{2}\rightarrow 2Fe_{2}O_{3}$

5.6 grams of Fe reacts with 3 moles of Oxygen.

To find:

Mass of Ferrous oxide produced.

Calculation:

$4Fe + 3O_{2}\rightarrow 2Fe_{2}O_{3} \\ (4 \: mol) \: (3 \: mol) \: \: (2 \: mol)$

Oxygen is given in excess of 3 moles.

But in given rectants, moles of Fe = 5.6/56 = 0.1

So, here the limiting reagent is Fe.

4 moles of Fe gives 2 moles of Fe2O3

=> 0.1 moles of Fe gives 2/4 × 0.1 = 0.05 moles.

Hence, let mass of Fe2O3 produced be m:

$\therefore \: m = moles \times molar \: mass$

$= > \: m = 0.05\times160$

$= > \: m = 0.5\times16$

$= > \: m = 8 \: gm$

So , 8 gm of ferrous oxide is produced.