How many liters of hydrogen gas at 754.3 mmHg and 723 K are required to react with 34.21 g of Iron (III) oxide?

Question

Small amounts of pure iron are made by treating Iron (III) oxide with hydrogen gas:Fe2O3 + H2 ---> Fe + H2O How many liters of hydrogen gas at 754.3 mmHg and 723 K are required to react with 34.21 g of Iron (III) oxide?

Answer 1

First, you must write a correctly balanced equation:

Fe₂O₃ + 3 H₂ ==> 2 Fe + 3H₂O

Find moles of iron oxide used, then moles of H₂ needed, and finally the volume of H₂.

moles Fe₂O₃ = 34.21 g x 1 mol/159.7 g = 0.2142 moles

moles H₂ needed = 0.2142 mol Fe₃O₂ x 3 mol H₂/mol Fe₃O₂ = 0.6426 moles H₂

Next, you can use PV = nRT to find volume of H₂

V = nRT/P = (0.6426)(0.0821)(723)/(754.3/760)

note: convert mm Hg to atm by dividing by 760 mm/arm

V = 381.5/0.9925 = 38.43 liters