Need help finding liters needed in a reduction reaction :) (Problem in desc)

Question

Iron is produced by reduction from its ore according to the reaction

 

Fe₂O_3(s) + 3 H_2(g) --> 2 Fe_(s) + 3 H₂O_(g)

 

How many liters of H₂ gas are required to completely react with 34.2 g of Fe₂O₃ at 754.3 mm Hg and a temperature of 723 K?

Struggling to wrap my mind around this for some reason, an explanation would mean everything :)

Answer 1

Fe₂O_3(s) + 3 H_2(g) --> 2 Fe_(s) + 3 H₂O_(g)

We are given the mass of Fe₂O₃ so from that, we can find the moles of Fe₂O₃. Once we have the moles of Fe₂O₃, we can use the stoichiometry of the equation to find the moles of H₂ needed to completely react. Then, using the ideal gas law, we can calculate the liters of H₂ equivalent to the number of moles just calculated. Right?

moles Fe₂O₃ present = 34.2 g x 1 mol / 159.7 g = 0.2142 moles

moles H₂ needed = 0.2142 mol Fe₂O₃ x 3 mol H₂ / 1 mol Fe₂O₃ = 0.6426 mols H₂ needed

Ideal gas law: PV = nRT

P = pressure in atm = 754.3 mm Hg x 1 atm / 760 mm Hg = 0.9925 atm

V = volume in L = ?

n = moles = 0.6426

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 723K

Solving for V we have...

V = nRT/P = (0.6426)(0.0821)(723) / 0.9925

V = 38.4 L

Did you follow all of that? Does it now make sense?