Question

How many grams of Oxygen gas must react with excess Iron in order to produce 44.8 L of Iron ( III) Oxide?

4 Fe + 3 O2 = 2 Fe2O3

Answer 1

$4 Fe + 3 O_2 = 2 Fe_2O_3$

44.8L of Fe₂O₃ = x mole

22.4L = 1 mole

x = 2 moles

From the reaction above,

3 moles of Oxygen produce 2 moles of Iron(III)Oxide

$\therefore$ 3 moles of Oxygen are produced

1 mole of Oxygen gas = 32g

3 moles of Oxygen gas is therefore 32(3)g = 96g

$\therefore$ 96 grams of Oxygen gas must react with excess Iron in order to produce 44.8L of Iron(III)Oxide.