Question

How many liters of oxygen gas, at standard

temperature and pressure, will react with 35.8 grams of
iron metal?
4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)​

Answer 1

Step-by-step explanation:

• For the balanced reaction:

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.

• Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

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

• Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

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.

• Finally, we can get the volume of oxygen using the information:

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.