How many liters of oxygen gas, at standardtemperature and pressure, will react with 35.8 grams ofiron metal? 4 fe (s) + 3 o2 (g) → 2 fe2o3 (s)

Explanation:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

Using cross multiplication:

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).

Using cross multiplication:

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

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